Let’s dive into the exciting world of derivatives, specifically focusing on trigonometric functions! When we talk about derivatives of trigonometric functions, we’re really looking at how these functions change with respect to a variable. This is a key concept in calculus, which you'll come across in exams like the SAT, ACT, PSAT, GRE, and even the MCAT.

In this article, we’ll break down the formulas for the derivatives of trigonometric functions and explore the process of differentiating them step by step. We’ll also cover some cool techniques, such as the First Principle of Derivatives, the Chain Rule, and the Quotient Rule. These are all super important, especially when you’re working through problems in exams like the AP Exams or the Accuplacer. With some practice, you'll see how these tools make finding derivatives easier and more fun. Plus, we’ll go through some solved examples so you can apply what you’ve learned and feel confident heading into your next exam!

Derivatives of Trigonometric Function

The derivative of a function is a concept in mathematics of a real variable that measures the sensitivity to change of the function value (output value) with respect to a change in its argument (input value). They are a part of differential calculus. There are various methods of differentiation. The derivative of a function of a single variable at a chosen input value, when it exists, is the slope of the tangent line to the graph of the function at that point. The derivatives are used to find solutions to differential equations.

The differentiation of trigonometric functions is the mathematical process of finding the derivative of a trigonometric function, or its rate of change with respect to a variable. All derivatives of circular trigonometric functions can be found from those of sin(x) and cos(x) by means of the quotient rule application of functions like tan(x) = sin(x)/cos(x).

Being able to calculate the derivatives of the sine and cosine functions will enable us to solve many problems such as finding the velocity and acceleration of simple harmonic motion.

Derivative Formula of Trigonometric Functions

The following table summarizes the derivatives of the six trigonometric functions, as well as their chain rule counterparts

Proof of Derivatives of Trigonometric Function

We already saw the derivative formula of trigonometric functions like sinx, cosx, tanx, cotx, secx and cosecx by using various differentiation rules like First Principle of Derivatives, Chain Rule and Quotient rule and Solved Examples. Now let’s see how we get the formula for all the Derivatives of Trigonometric Function.

Derivative of sinx by the First Principle

Derivative of sin x by the first principle refers to using algebra to find a general expression for the slope of a curve. It is also known as the delta method. The derivative is a measure of the instantaneous rate of change.

If f(x) = sinx , find f’(x)

\(\begin{matrix}
f’(x)={dy\over{dx}}=\lim _{h{\rightarrow}0}{f(x+h)–f(x)\over{h}}
f(x)=sinx\\
f(x+h)=sin(x+h)\\
f(x+h)–f(x)= sin(x+h) – sin(x) = sinxcosh + cosxsinh – sinx\\
= sinx(cosh-1) + cosxsinh\\
{f(x+h) – f(x)\over{h}}={ sinx(cosh-1) + cosxsinh\over{h}}\\
\lim _{h{\rightarrow}0}{f(x+h) –f(x)\over{h}} = \lim _{h{\rightarrow}0} { sinx(cosh-1) + cosxsinh\over{h}}\\
\lim _{h{\rightarrow}0}{f(x+h) –f(x)\over{h}} = \lim _{h{\rightarrow}0} {sinx(cosh-1)\over{h}} + \lim _{h{\rightarrow}0} {cosxsinh\over{h}}\\
= sinx \lim _{h{\rightarrow}0} {(cosh-1)\over{h}} + cosx \lim _{h{\rightarrow}0} {sinh\over{h}}\\
\text{Put h = 0 in first limit}\\
sinx \lim _{h{\rightarrow}0} {(cosh-1)\over{h}} = sinx\times0 = 0\\
\text{Using L’ Hospitals Rule on Second Limit}\\
\lim _{h{\rightarrow}0}{f(x+h) –f(x)\over{h}} = cosx \lim _{h{\rightarrow}0} {{d\over{dh}}sinh\over{{d\over{dh}}h}}\\
\lim _{h{\rightarrow}0}{f(x+h) –f(x)\over{h}} = cosx \lim _{h{\rightarrow}0} {cosh\over{1}}\\
\lim _{h{\rightarrow}0}{f(x+h) –f(x)\over{h}} = cosx \times1 = cosx\\
f’(x)={dy\over{dx}} = {d(sinx)\over{dx}} = cosx
\end{matrix}\)

Alternatively

\(\begin{matrix}\
f’(x)={dy\over{dx}}=\lim _{h{\rightarrow}0}{f(x+h)–f(x)\over{h}}
f(x)=sinx\\
f(x+h)=sin(x+h)\\
f(x+h)–f(x)= sin(x+h) – sin(x) = {2cos({x+h+x\over{2}})sin({x+h-x\over{2}})\over{h}}\\
\lim _{h{\rightarrow}0}{f(x+h) –f(x)\over{h}} = \lim _{h{\rightarrow}0} {2cos({x+h+x\over{2}})sin({x+h-x\over{2}})\over{h}}\\
\lim _{h{\rightarrow}0}{f(x+h) –f(x)\over{h}} = \lim _{h{\rightarrow}0} 2cos({x+h+x\over{2}}){sin({x+h-x\over{2}})\over{h}}\\
\lim _{h{\rightarrow}0}{f(x+h) –f(x)\over{h}} = \lim _{h{\rightarrow}0}2cos({x+h+x\over{2}}){sin({x+h-x\over{2}})\over{{h\over{2}}}}\\
\lim _{h{\rightarrow}0}{f(x+h) –f(x)\over{h}} = \lim _{h{\rightarrow}0} 2cos({x+h+x\over{2}})\times1\\
{\because}\lim _{h{\rightarrow}0}{sin({h\over{2}})\over{{h\over{2}}}} = 1\\
\lim _{h{\rightarrow}0}{f(x+h) –f(x)\over{h}} = \lim _{h{\rightarrow}0} 2cos({x+h+x\over{2}}) = cosx\\
f’(x)={dy\over{dx}} = {d(sinx)\over{dx}} = cosx
\end{matrix}\)

Derivative of sinx by the Chain Rule

Chain Rule helps us differentiate composite functions with the number of functions that make up the composition determining how many differentiation steps are necessary.

If f(x) = sinx , find f’(x)

\(\begin{matrix}
f(x) = sinx = cos({\pi\over{2}} -x )\\
\text{ Using chain rule, }\\
f’(x) = – sin({\pi\over{2}} -x) · {d\over{dx}} ({\pi\over{2}} -x)\\
= – sin({\pi\over{2}} -x) · (-1)\\
= sin({\pi\over{2}} -x)\\
= cosx
\end{matrix}\)

Derivative of sinx by the Quotient Rule

A special rule, the quotient rule, exists for differentiating quotients of two functions. Functions often come as quotients, by which we mean one function divided by another function. There is a formula we can use to differentiate a quotient – it is called the quotient rule.

If f(x) = sinx , find f’(x)

\(\begin{matrix}
f(x) = sinx = {1\over{cosecx}}\\
\text{ Using quotient rule, }\\
f’(x) = {cosecx{d\over{dx}}(1) – 1.{d\over{dx}}cosecx\over{cosec^2x}}\\
= {cosecxcotx\over{cosec^2x}}\\
= {cotx\over{cosecx}}\\
= {{cosx\over{sinx}}\over{{1\over{sinx}}}}\\
= cosx
\end{matrix}\)

Derivative of cosx by the First Principle

Derivative of cos x by the first principle refers to using algebra to find a general expression for the slope of a curve. It is also known as the delta method. The derivative is a measure of the instantaneous rate of change.

If f(x) = cosx , find f’(x)

\(\begin{matrix}\
f’(x)={dy\over{dx}}=\lim _{h{\rightarrow}0}{f(x+h)–f(x)\over{h}}
f(x)=cosx\\
f(x+h)=cos(x+h)\\
f(x+h)–f(x)= cos(x+h) – cos(x) = cosxcosh – sinxsinh – cosx\\
= cosx(cosh-1) – sinxsinh\\
{f(x+h) – f(x)\over{h}}={ cosx(cosh-1) – sinxsinh\over{h}}\\
\lim _{h{\rightarrow}0}{f(x+h) –f(x)\over{h}} = \lim _{h{\rightarrow}0} { cosx(cosh-1) – sinxsinh\over{h}}\\
\lim _{h{\rightarrow}0}{f(x+h) –f(x)\over{h}} = \lim _{h{\rightarrow}0} {cosx(cosh-1)\over{h}} – \lim _{h{\rightarrow}0} {sinxsinh\over{h}}\\
= cosx \lim _{h{\rightarrow}0} {(cosh-1)\over{h}} – sinx \lim _{h{\rightarrow}0} {sinh\over{h}}\\
\text{Put h = 0 in first limit}\\
cosx \lim _{h{\rightarrow}0} {(cosh-1)\over{h}} = cosx\times0 = 0\\
\text{Using L’ Hospitals Rule on Second Limit}\\
\lim _{h{\rightarrow}0}{f(x+h) –f(x)\over{h}} = -sinx \lim _{h{\rightarrow}0} {{d\over{dh}}sinh\over{{d\over{dh}}h}}\\
\lim _{h{\rightarrow}0}{f(x+h) –f(x)\over{h}} = -sinx \lim _{h{\rightarrow}0} {cosh\over{1}}\\
\lim _{h{\rightarrow}0}{f(x+h) –f(x)\over{h}} = -sinx \times1 = -sinx\\
f’(x)={dy\over{dx}} = {d(cosx)\over{dx}} = -sinx
\end{matrix}\)

Alternatively,

\(\begin{matrix}\
f’(x)={dy\over{dx}}=\lim _{h{\rightarrow}0}{f(x+h)–f(x)\over{h}}
f(x)=cosx\\
f(x+h)=cos(x+h)\\
f(x+h)–f(x)= cos(x+h) – cos(x) = {-2sin({x+h+x\over{2}})sin({x+h-x\over{2}})\over{h}}\\
\lim _{h{\rightarrow}0}{f(x+h) –f(x)\over{h}} = \lim _{h{\rightarrow}0} {-2sin({2x+h\over{2}})sin({h\over{2}})\over{h}}\\
\lim _{h{\rightarrow}0}{f(x+h) –f(x)\over{h}} = \lim _{h{\rightarrow}0} -2cos(x+{h\over{2}}){sin({h\over{2}})\over{h}}\\
\lim _{h{\rightarrow}0}{f(x+h) –f(x)\over{h}} = \lim _{h{\rightarrow}0}-2sin(x+{h\over{2}}){sin({h\over{2}})\over{{h\over{2}}}}\\
\lim _{h{\rightarrow}0}{f(x+h) –f(x)\over{h}} = \lim _{h{\rightarrow}0} -2sin(x+{h\over{2}})\times1\\
{\because}\lim _{h{\rightarrow}0}{sin({h\over{2}})\over{{h\over{2}}}} = 1\\
\lim _{h{\rightarrow}0}{f(x+h) –f(x)\over{h}} = \lim _{h{\rightarrow}0} -2sin(x+{h\over{2}}) = -sinx\\
f’(x)={dy\over{dx}} = {d(sinx)\over{dx}} = -sinx
\end{matrix}\)

Derivative of cosx by the Chain Rule

Chain Rule helps us differentiate composite functions with the number of functions that make up the composition determining how many differentiation steps are necessary.

If f(x) = cosx , find f’(x)

\(\begin{matrix}
f(x) = cosx = sin({\pi\over{2}} -x )\\
\text{ Using chain rule, }\\
f’(x) = cos({\pi\over{2}} -x) · {d\over{dx}} ({\pi\over{2}} -x)\\
= cos({\pi\over{2}} -x) · (-1)\\
= -cos({\pi\over{2}} -x)\\
= -sinx
\end{matrix}\)

Derivative of cosx by the Quotient Rule

A special rule, the quotient rule, exists for differentiating quotients of two functions. Functions often come as quotients, by which we mean one function divided by another function. There is a formula we can use to differentiate a quotient – it is called the quotient rule.

If f(x) = cosx , find f’(x)

\(\begin{matrix}
f(x) = cosx = {1\over{secx}}\\
\text{ Using quotient rule, }\\
f’(x) = {secx{d\over{dx}}(1) – 1.{d\over{dx}}secx\over{sec^2x}}\\
= -{secxtanx\over{sec^2x}}\\
= -{tanx\over{secx}}\\
= -{{sinx\over{cosx}}\over{{1\over{cosx}}}}\\
= -sinx
\end{matrix}\)

Derivative of tanx by the First Principle

Derivative by the first principle refers to using algebra to find a general expression for the slope of a curve. It is also known as the delta method. The derivative is a measure of the instantaneous rate of change.

If f(x) = tanx , find f’(x)

\(\begin{matrix}\
f’(x)={dy\over{dx}}=\lim _{h{\rightarrow}0}{f(x+h)–f(x)\over{h}}
f(x)=tanx\\
f(x+h)=tan(x+h)\\
f(x+h)–f(x)= tan(x+h) – tan(x) = {sin(x+h)\over{cos(x+h)}} – {sin(x)\over{cos(x)}}\\
{f(x+h) – f(x)\over{h}}={ {sin(x+h)\over{cos(x+h)}} – {sin(x)\over{cos(x)}}\over{h}}\\
\lim _{h{\rightarrow}0}{f(x+h) –f(x)\over{h}} = \lim _{h{\rightarrow}0} { {sin(x+h)\over{cos(x+h)}} – {sin(x)\over{cos(x)}}\over{h}}\\
\lim _{h{\rightarrow}0}{f(x+h) –f(x)\over{h}} = \lim _{h{\rightarrow}0} {cosxsin(x+h) – sinxcos(x+h)\over{hcosxcos(x+h)}}\\
\lim _{h{\rightarrow}0}{f(x+h) –f(x)\over{h}} = \lim _{h{\rightarrow}0} {{sin(2x+h)+sinh\over{2}} – {sin(2x+h)-sinh\over{2}}\over{hcosxcos(x+h)}}\\
\lim _{h{\rightarrow}0}{f(x+h) –f(x)\over{h}} = \lim _{h{\rightarrow}0} {sinh\over{hcosxcos(x+h)}}\\
\lim _{h{\rightarrow}0}{f(x+h) –f(x)\over{h}} = \lim _{h{\rightarrow}0} {sinh\over{h}}\\
\lim _{h{\rightarrow}0}{f(x+h) –f(x)\over{h}} = \lim _{h{\rightarrow}0} {1\over{cosxcos(x+h)}}\\
=1\times{1\over{cosx\times{cosx}}}
={1\over{cos^2x}}
={sec^2x}
\lim _{h{\rightarrow}0}{f(x+h) –f(x)\over{h}} = {sec^2x}\\
f’(x)={dy\over{dx}} = {d(tanx)\over{dx}} = {sec^2x}
\end{matrix}\)

Derivative of tanx by the Chain Rule

Chain Rule helps us differentiate composite functions with the number of functions that make up the composition determining how many differentiation steps are necessary.

If f(x) = tanx , find f’(x)

\(\begin{matrix}
f(x) = tanx = {sinx\over{cosx}}\\
\text{ Using chain rule, }\\
f’(x) = {sinx{d\over{dx}}(cosx) – cosx{d\over{dx}}sinx\over{cos^2x}}\\
= {sinx.sinx – cosx(-cosx)\over{cos^2x}}\\
= {sin^2x + cos^2x\over{cos^2x}}\\
= {1\over{cos^2x}}\\
= sec^2x
\end{matrix}\)

Derivative of tanx by the Quotient Rule

A special rule, the quotient rule, exists for differentiating quotients of two functions. Functions often come as quotients, by which we mean one function divided by another function. There is a formula we can use to differentiate a quotient – it is called the quotient rule.

If f(x) = tanx , find f’(x)

\(\begin{matrix}
f(x) = tanx = {1\over{cotx}}\\
\text{ Using quotient rule, }\\
f’(x) = {cotx{d\over{dx}}(1) – 1.{d\over{dx}}cotx\over{cot^2x}}\\
= {cosec^2x\over{cot^2x}}\\
= {{sin^2x\over{cos^2x}}\over{{1\over{sin^2x}}}}\\
= {1\over{cos^2x}}\\
= sec^2x
\end{matrix}\)

Derivative of cotx by the First Principle

Derivative by the first principle refers to using algebra to find a general expression for the slope of a curve. It is also known as the delta method. The derivative is a measure of the instantaneous rate of change.

If f(x) = cotx , find f’(x)

\(\begin{matrix}\
f’(x)={dy\over{dx}}=\lim _{h{\rightarrow}0}{f(x+h)–f(x)\over{h}}
f(x)=cotx\\
f(x+h)=cot(x+h)\\
f(x+h)–f(x)= cot(x+h) – cot(x) = {cos(x+h)\over{sin(x+h)}} – {cos(x)\over{sin(x)}}\\
{f(x+h) – f(x)\over{h}}={{cos(x+h)\over{sin(x+h)}} – {cos(x)\over{sin(x)}}\over{h}}\\
\lim _{h{\rightarrow}0}{f(x+h) –f(x)\over{h}} = \lim _{h{\rightarrow}0} {{cos(x+h)\over{sin(x+h)}} – {cos(x)\over{sin(x)}}\over{h}}\\
\lim _{h{\rightarrow}0}{f(x+h) –f(x)\over{h}} = \lim _{h{\rightarrow}0} {sinxcos(x+h) – cosxsin(x+h)\over{hsinxsin(x+h)}}\\
\lim _{h{\rightarrow}0}{f(x+h) –f(x)\over{h}} = \lim _{h{\rightarrow}0} {{cos(2x+h)+cosh\over{2}} – {cos(2x+h)-cosh\over{2}}\over{hsinxsin(x+h)}}\\
\lim _{h{\rightarrow}0}{f(x+h) –f(x)\over{h}} = \lim _{h{\rightarrow}0} {-cosh\over{hsinxsin(x+h)}}\\
\lim _{h{\rightarrow}0}{f(x+h) –f(x)\over{h}} = \lim _{h{\rightarrow}0} {-cosh\over{h}}\\
\lim _{h{\rightarrow}0}{f(x+h) –f(x)\over{h}} = \lim _{h{\rightarrow}0} {-1\over{sinxsin(x+h)}}\\
=(-1)\times{1\over{sinx\times{sinx}}}\\
={-1\over{cosec^2x}}\\
={-cosec^2x}\\
\lim _{h{\rightarrow}0}{f(x+h) –f(x)\over{h}} = {-cosec^2x}\\
f’(x)={dy\over{dx}} = {d(cotx)\over{dx}} = {-cosec^2x}
\end{matrix}\)

Derivative of cotx by the Chain Rule

Chain Rule helps us differentiate composite functions with the number of functions that make up the composition determining how many differentiation steps are necessary.

If f(x) = cotx , find f’(x)

\(\begin{matrix}
f(x) = tanx = {cosx\over{sinx}}\\
\text{ Using chain rule, }\\
f’(x) = {cosx{d\over{dx}}(sinx) – sinx{d\over{dx}}cosx\over{sin^2x}}\\
= {-cosx.cosx – sinx(-sinx)\over{sin^2x}}\\
= -{cos^2x + sin^2x\over{sin^2x}}\\
= -{1\over{sin^2x}}\\
= -cosec^2x
\end{matrix}\)

Derivative of cotx by the Quotient Rule

A special rule, the quotient rule, exists for differentiating quotients of two functions. Functions often come as quotients, by which we mean one function divided by another function. There is a formula we can use to differentiate a quotient – it is called the quotient rule.

If f(x) = tanx , find f’(x)

\(\begin{matrix}
f(x) = cotx = {1\over{tanx}} = tan^{-1}x\\
\text{ Using quotient rule, }\\
f’(x) =(-1)tan^{-2}xsec^2x\\
= -{sec^2x\over{tan^2x}}\\
= -{{cos^2x\over{sin^2x}}\over{{1\over{cos^2x}}}}\\
= -{1\over{sin2x}}\\
= -cosec^2x
\end{matrix}\)

Derivative of secx by the First Principle

Derivative by the first principle refers to using algebra to find a general expression for the slope of a curve. It is also known as the delta method. The derivative is a measure of the instantaneous rate of change.

If f(x) = secx , find f’(x)

\(\begin{matrix}\
f’(x)={dy\over{dx}}=\lim _{h{\rightarrow}0}{f(x+h)–f(x)\over{h}}
f(x)=secx\\
f(x+h)=sec(x+h)\\
f(x+h)–f(x)= sec(x+h) – sec(x) = {1\over{(cos (x + h)}} – {1\over{cos x}}\\
= {cos x – cos(x + h)\over{cos x cos(x + h)}}\\
{f(x+h) – f(x)\over{h}}={{cos x – cos(x + h)\over{cos x cos(x + h)}}\over{h}}\\
\text{ By sum to product formulas, cos A – cos B = -2 sin (A+B)/2 sin (A-B)/2. }\\
{f(x+h) – f(x)\over{h}}={1\over{cos x }}{[- 2 sin (x + x + h)/2 sin (x – x – h)/2] / [cos(x + h)]\over{h}}\\
{f(x+h) – f(x)\over{h}}={1\over{cos x }}{ [- 2 sin (2x + h)/2 sin (- h)/2] / [cos(x + h)]\over{h}}\\
\lim _{h{\rightarrow}0}{f(x+h) –f(x)\over{h}} = \lim _{h{\rightarrow}0} {1\over{cos x }}{ [- 2 sin (2x + h)/2 sin (- h)/2] / [cos(x + h)]\over{h}}\\
\lim _{h{\rightarrow}0}{f(x+h) –f(x)\over{h}} = {1\over{cos x }}\lim _{h{\rightarrow}0}{ [- 2 sin (2x + h)/2 sin (- h)/2] / [cos(x + h)]\over{h}}\\
\text{ Multiply and divide by h/2,}\\
\lim _{h{\rightarrow}0}{f(x+h) –f(x)\over{h}} = {1\over{cos x }}\lim _{h{\rightarrow}0}({1\over{h}}) ({h\over{2}}) [- 2 sin (2x + h)/2 sin (- h/2) / (h/2)] / [cos(x + h)]\\
\text{ When h → 0, we have h/2 → 0. So }\\
\lim _{h{\rightarrow}0}{f(x+h) –f(x)\over{h}} = {1\over{cos x }}\lim _{{h\over{2}}{\rightarrow}0}{sin(h/2)\over{(h/2)}}\lim _{h{\rightarrow}0}(sin(2x + h)/2)/cos(x + h)\\
={1\over{cos x}}{sin x\over{cos x}}\\
=secxtanx\\
f’(x)={dy\over{dx}} = {d(secx)\over{dx}} = secxtanx
\end{matrix}\)

Derivative of secx by the Chain Rule

Chain Rule helps us differentiate composite functions with the number of functions that make up the composition determining how many differentiation steps are necessary.

If f(x) = secx , find f’(x)

\(\begin{matrix}
f(x) = secx = {1\over{cos x}}\\
\text{ Using chain rule, }\\
f’(x) = {cosx{d\over{dx}}(1) – 1.{d\over{dx}}cosx\over{cos^2x}}\\
= {sinx\over{cos^2x}}\\
= {{sinx\over{cosx}}\over{{1\over{cosx}}}}\\
=secxtanx
\end{matrix}\)

Derivative of secx by the Quotient Rule

A special rule, the quotient rule, exists for differentiating quotients of two functions. Functions often come as quotients, by which we mean one function divided by another function. There is a formula we can use to differentiate a quotient – it is called the quotient rule.

If f(x) = secx , find f’(x)

\(\begin{matrix}
f(x) = secx = {1\over{cosx}} = cos^{-1}x\\
\text{ Using quotient rule, }\\
f’(x) =(-1)cos^{-2}x(-sinx)\\
= {(sin x)\over{cos^2x}}\\
= {{sinx\over{cosx}}\over{{1\over{cosx}}}}\\
= secxtanx
\end{matrix}\)

Derivative of cosecx by the First Principle

Derivative by the first principle refers to using algebra to find a general expression for the slope of a curve. It is also known as the delta method. The derivative is a measure of the instantaneous rate of change.

If f(x) = cosecx , find f’(x)

\(\begin{matrix}\
f’(x)={dy\over{dx}}=\lim _{h{\rightarrow}0}{f(x+h)–f(x)\over{h}}
f(x)=cosecx\\
f(x+h)=cosec(x+h)\\
f(x+h)–f(x)= cosec(x+h) – cosec(x) = {1\over{(sin(x + h)}} – {1\over{sinx}}\\
= {sinx – sin(x + h)\over{sin x sin(x + h)}}\\
{f(x+h) – f(x)\over{h}}={{sin x – sin(x + h)\over{sin x sin(x + h)}}\over{h}}\\
{f(x+h) – f(x)\over{h}}={sin x – sin(x + h)\over{h{sin x sin(x + h)}}}\\
{f(x+h) – f(x)\over{h}}={-{sin(x + h) – sinx\over{h}}\times{1\over{{sin x sin(x + h)}}}}\\
\lim _{h{\rightarrow}0}{f(x+h) –f(x)\over{h}} = \lim _{h{\rightarrow}0}[ {-{sin(x + h) – sinx\over{h}}\times{1\over{{sin x sin(x + h)}}}}]\\
\lim _{h{\rightarrow}0}{f(x+h) –f(x)\over{h}} = \lim _{h{\rightarrow}0}{-{sin(x + h) – sinx\over{h}}\times{1\over{ \lim _{h{\rightarrow}0}{sin x sin(x + h)}}}}]\\
=-cosx\times{1\over{sin^2x}}\\
=-{cosx\over{sinx}}\times{1\over{sinx}}\\
=-cotxcosecx\\
f’(x)={dy\over{dx}} = {d(cosecx)\over{dx}} = -cotxcosecx
\end{matrix}\)

Derivative of cosecx by the Chain Rule

Chain Rule helps us differentiate composite functions with the number of functions that make up the composition determining how many differentiation steps are necessary.

If f(x) = cosecx , find f’(x)

\(\begin{matrix}
f(x) = cosecx = {1\over{sin x}}\\
\text{ Using chain rule, }\\
f’(x) = {sinx{d\over{dx}}(1) – 1.{d\over{dx}}sinx\over{sin^2x}}\\
= -{cosx\over{sin^2x}}\\
=-{cosx\over{sinx}}\times{1\over{sinx}}\\
=-cotxcosecx
\end{matrix}\)

Derivative of cosecx by the Quotient Rule

A special rule, the quotient rule, exists for differentiating quotients of two functions. Functions often come as quotients, by which we mean one function divided by another function. There is a formula we can use to differentiate a quotient – it is called the quotient rule.

If f(x) = cosecx , find f’(x)

\(\begin{matrix}
f(x) = cosecx = {1\over{sinx}} = sin^{-1}x\\
\text{ Using quotient rule, }\\
f’(x) =(-1)sin^{-2}x(cosx)\\
= {-(cosx)\over{sin^2x}}\\
= {-{cosx\over{sinx}}\over{{1\over{sinx}}}}\\
=-cotxcosecx
\end{matrix}\)

Real-World Applications of Trigonometric Functions

Trigonometric function derivatives find extensive applications in real-world scenarios, especially in fields like physics, engineering, and computer science. One prominent application is in the analysis of periodic motion. For instance, when studying the oscillations of a pendulum or the vibrations of a guitar string, derivatives of trigonometric functions are employed to determine the amplitude, frequency, and phase of these periodic motions. In structural engineering, understanding the derivatives of trigonometric functions is crucial for predicting and mitigating vibrations in buildings and bridges.
Moreover, these derivatives play a vital role in signal processing and telecommunications. The modulation of signals, essential for transmitting information, relies on trigonometric functions and their derivatives. In electrical engineering, the derivatives of trigonometric functions contribute to the design of circuits and systems, ensuring efficient energy transfer and signal integrity.

Advanced Techniques in Trigonometric Differentiation

Moving beyond elementary applications, advanced techniques in trigonometric differentiation provide a deeper understanding of these functions and their derivatives. Implicit differentiation is a powerful method applied when functions are present implicitly, allowing for the differentiation of complex relationships involving trigonometric functions. This technique is valuable in scenarios where explicit representations are challenging to establish.
Parametric differentiation is another advanced approach that is frequently employed in physics and engineering. When dealing with curves defined parametrically, such as the motion of a projectile, parametric differentiation allows for the precise calculation of derivatives concerning time, enabling the determination of velocity and acceleration.
The concept of logarithmic differentiation offers yet another advanced tool. By taking the natural logarithm of both sides of an equation, complex expressions involving products and quotients simplify, making the application of trigonometric differentiation rules more manageable. This technique is particularly handy in scenarios where straightforward differentiation may lead to cumbersome algebraic manipulations.

Solved Examples on Derivatives of Trigonometric Function

Solved Example 1 : Find the derivative of \(f(x) = 5x^3sinx\)

Using the product rule, we have

\(\begin{matrix}
f'(x) = sinx{d\over{dx}}(5x^3) + 5x^3{d\over{dx}}(sinx)\\
= sinx15x^2 + 5x^3cosx\\
\text{ After simplifying, we obtain }\\
f′(x) = 15x^2sinx + 5x^3cosx.
\end{matrix}\)

Solved Example 2 : Solve the following \(f(x) = {cosx\over{4x^2}}᠎\)

A.By applying the quotient rule, we have

\(\begin{matrix}
f′(x) = {(−sinx)4x^2 − 8x(cosx)\over{(4x^2)^2}}\\
\text{ Simplifying, we obtain }\\
f′(x) = {−4x^2sinx − 8xcosx\over{16x^4}}\\
f’(x) = {−xsinx−2cosx\over{4x^3}}
\end{matrix}​᠎​\)

1. Use the Product rule to differentiate the given function, ᠎\( y = 5xsinx + 4x^2cosx ᠎\)

\(\begin{matrix}
\text{ Differentiating using Product Rule, }\\
y’= 5sinx + 5xcosx + 8xcosx – 4x^2sinx\\
y’= 5sinx + 13xcosx – 4x^2sinx
\end{matrix}​᠎​\)

Questions to Ace Your Exams

Q1.What is the derivative of trigonometric functions?

The derivatives of trigonometric functions are sine (sin) and cosine (cos). They adhere to distinct rules. The derivative is cos(x) for sin(x), and it becomes -sin(x) for cos(x). These patterns cyclically repeat every four derivatives, forming a foundational aspect of trigonometric calculus. When calculating derivatives, employ these rules and incorporate the chain rule for composite functions involving trigonometry. Developing a comprehensive understanding of these fundamental principles is imperative for navigating and solving derivative problems within the realm of trigonometric functions. Mastery of these rules enables a more streamlined and efficient approach to derivative computations in trigonometry, facilitating problem-solving across various mathematical applications.

Q2.How to find the derivative of a trigonometric function? How to find the nth derivative of a trigonometric function? How to solve derivatives of trigonometric functions?

When seeking the derivatives of trigonometric functions, follow specific rules: sin(x) differentiates to cos(x) and cos(x) to -sin(x). Employ the chain rule for compositions involving trigonometric functions. For the nth derivative, recognize the cyclical pattern repeating every four iterations, streamlining calculations. To solve derivatives of trigonometric functions effectively, leverage trigonometric identities to simplify expressions. Apply the product, quotient, and chain rules as needed for accurate solutions. Developing a systematic approach and a deep understanding of trigonometric identities is crucial. This knowledge not only facilitates finding derivatives but also ensures proficiency in solving complex trigonometric derivative problems. Mastery of these techniques empowers a comprehensive and efficient approach to navigating the intricacies of trigonometric calculus.

Q3.What is the derivative of inverse trigonometric functions?

The derivatives of inverse trigonometric functions like arcsin(x) and arctan(x) have specific formulas crucial in calculus. The derivative for arcsin(x) is 1/√(1-x^2). It emphasizes the reciprocal of the square root of the difference between 1 and the square of the variable. The derivative of arctan(x) is 1/(1 x^2). It highlights the importance of the reciprocal of the sum of 1 and the square of the variable. These derivative expressions are fundamental tools for calculating rates of change and slopes in problems involving inverse trigonometric functions, contributing to a deeper understanding of calculus applications and mathematical modeling. Mastery of these formulas enhances proficiency in tackling diverse mathematical scenarios involving inverse trigonometric functions.

Q4.How to find derivatives of higher powers of trigonometric functions?

When finding derivatives of higher powers of trigonometric functions, apply the power rule and utilize trigonometric derivative rules. For example, differentiating sin^2(x) involves the power rule (2sin(x)cos(x)) and trigonometric identities for simplification. In cases like sin^3(x), the chain rule is crucial, requiring the derivative of sin(x) multiplied by the square of sin(x). A solid grasp of these rules, combined with familiarity with trigonometric identities, is essential. Regular practice in simplifying expressions using these techniques enhances efficiency in calculating derivatives of higher powers of trigonometric functions, contributing to a more nuanced understanding of calculus within the realm of trigonometry.

Conclusion

In conclusion, mastering the derivatives of trigonometric functions is key to solving many calculus problems, especially in exams like the SAT, ACT, and AP. By using techniques like the First Principle of Derivatives, the Chain Rule, and the Quotient Rule, you'll be equipped to differentiate functions like sin(x), cos(x), tan(x), and more. Practice with solved examples will help solidify your understanding, making it easier to tackle any related questions in your exams.