Evaluation of derivatives MCQ Quiz in मल्याळम - Objective Question with Answer for Evaluation of derivatives - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

• \(\rm d\over dx\)xn = nxn-1
• \(\rm d\over dx\)sin x = cos x
• \(\rm d\over dx\)cos x = -sin x
• \(\rm d\over dx\)ex = ex
• \(\rm d\over dx\)ln x = \(\rm1\over x\)
• \(\rm d\over dx\)(ax + b) = a
• \(\rm d\over dx\)tan x = sec2 x

Chain Rule: If y is a function of u and u is a function of x

• \(\rm {dy\over dx} = {dy\over du}\times {du\over dx}\)

Calculation:

f(x) = \(\rm \left(x+{1\over x}\right)^2 \)

f(x) = \(\rm \left(x^2+{1\over x^2}+2\right) \)

f(x) = x² + x^-2 + 2

Differentiating w.r.t. x

f'(x) = \(\rm d\over dx\)f(x)

f'(x) = \(\rm d\over dx\)(x2 + x-2 + 2)

f'(x) = \(\rm d\over dx\)x² + \(\rm d\over dx\)x^-2 + \(\rm d\over dx\)2

f'(x) = (2x) + (-2x^-3) + 0

f'(x) = \(\boldsymbol{\rm 2\left(x - {1\over x^3}\right)}\)

Concept:

If x = f(t) and y = g(t)

then \(\frac{{dy}}{{dx}}\) = \({{dy \over dt}\over {dx \over dt}}\)

Calculation:

Given, y = sin(2sin-1 x),

let x = sin θ ⇒ y = sin (2sin-1(sin θ)),

⇒ y = sin 2θ 

Differentiating x and y with respect to θ,

⇒ \(dx \over dθ\) = cos θ and \(dy \over dθ\) = 2cos 2θ 

⇒ \(\frac{{dy}}{{dx}}\) = \({{dy \over dθ}\over {dx \over dθ}} = {2 cos 2θ \over cos θ}\) 

Now sin θ = x ⇒ cos θ = \(\sqrt{1 - x^2}\)

and cos 2θ = 1 - 2sin²θ = 1 - 2x²

⇒ \(\frac{{dy}}{{dx}} = 2{\cos2\theta \over \cos \theta} \) = \(\frac{{2 - 4{x^2}}}{{\sqrt {1 - {x^2}} }}\)

∴ The correct answer is option (1).

Concept:

Formula:

log mⁿ = n log m

\(\rm \frac{d(uv)}{dx} = v\frac{du}{dx}+u\frac{dv}{dx}\)

\(\rm \frac{d\log x}{dx} = \frac{1}{x}\)

Calculation:

Let y = x^x

Taking log both sides, we get

⇒ log y = log x^x

⇒ log y = x log x            (∵ log mⁿ = n log m)

Differentiating with respect to x, we get

\(\rm \Rightarrow \frac{1}{y}\frac{dy}{dx} = \log x \frac{dx}{dx} + x \frac{d\log x}{dx}\\\Rightarrow \frac{dy}{dx} = y \left(\log x + x \times \frac{1}{x} \right )\\\therefore \frac{dy}{dx} = x^x (1+\log x)\)

Concept-

sin 2x = 2sin x cos x.

Calculation-

As sin x = sin 2x

⇒ sin 2x - sin x = 0

⇒ 2 sin x cos x - sin x = 0

⇒ sin x (2cos x - 1) = 0

So either sin x = 0 , in interval [0, 2π] when x = 0, π, 2π 

or 2cos x -1 = 0, i.e cos x = \( {1} \over {2}\) in interval [0,2π] when x = \(\frac{ π } {3} ,\frac { 5π } {3}\)

∴ total values of x in interval [0, 2π] is 5.

Concept used:

Trigonometry formula

sin 2x = 2sin x cos x

Calculation:

y = \(\rm {\rm{\;}}\frac{{\left( {sinx - cosx} \right)}}{{sin2x}}\)

y = \(\rm \frac{{\left( {sinx - cosx} \right)}}{{2.sinx.cosx}}\)

⇒ y = \(\rm \frac{{sinx}}{{2.sinx.cosx}} - \;\frac{{cosx}}{{2.sinx.cosx}}\)

⇒ y = \(\rm \left( {\frac{1}{{2cosx}}} \right)\; - \left( {\frac{1}{{2sinx}}} \right)\)

⇒ y = \(\frac{1}{2}\) (secx - cosecx)

Differentiate both sides, we get

⇒ \(\rm \frac{{dy}}{{dx}} = \frac{1}{2}\left[ {\frac{{d\left( {secx} \right)}}{{dx}} - \frac{{d\left( {cosecx} \right)}}{{dx}}} \right]\)

⇒ \(\rm \frac{{dy}}{{dx}} = \frac{1}{2}\left( {secx.tanx + cosecx.cotx} \right)\)

Given:
\(\rm y=x^{\sec^2x}\times\frac{1}{x^{\tan^2x}}\)

Concept:

Use trigonometric identity \(\rm \sec^2x-\tan^2x=1\)

And use derivative rule \(\rm \frac{d}{dx}x^n=n x^{n-1}\)

Calculation:

\(\rm y=x^{\sec^2x}\times\frac{1}{x^{\tan^2x}}\)

\(\rm \implies y=x^{\sec^2x}\times x^{-\tan^2x}\)

\(\rm \implies y=x^{\sec^2x-\tan^2x}\)

\(\rm \implies y=x\)

Now differentiate with respect to \(\rm x\)

we get

\(\rm \frac{dy}{dx}=1\)

Hence the option (4) is correct.

Given:

The derivative of \(\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)\) with respect to tan-1x is:

Concept:

The derivative of f(x) with respect to g(x) is \(\rm \frac{f'(x)}{g'(x)}\)

Calculation:

Let \(f(x)=\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)\) and g(x) = tan-1x

Put x = tanθ in f(x) then

\(\rm f(x)=\tan^{-1}\left(\frac{sec\theta-1}{tan\theta}\right)\)

\(\rm f(x)=\tan^{-1}\left(\frac{1-cos\theta}{sin\theta}\right)\)

\(\rm f(x)=\tan^{-1}\left(\frac{sin^2\frac{\theta}{2}}{2sin\frac{\theta}{2}cos\frac{\theta}{2}}\right)\)

\(\rm f(x)=\tan^{-1}\left(\tan\frac{\theta}{2}\right)\)

\(\rm f(x)=\frac{\theta}{2}\)

\(\rm f(x)=\frac12tan^{-1}x\)

On differentiating .

\(\rm f'(x)=\frac{1}{2}\cdot\frac{1}{1+x^2}\)

we have g(x) = tan-1x

On differentiating 

\(\rm g'(x)=\frac{1}{1+x^2}\)

Now the derivative of f(x) with respect to g(x) is

\(\rm \frac{f'(x)}{g'(x)}=\frac{\frac{1}{2}\cdot\frac{1}{1+x^2}}{\frac{1}{1+x^2}}=\frac{1}{2}\)

Hence the option (2) is correct.

Calculation:

Given:

ex + ey = ex + y = e^xe^y

e^-y + e^-x = 1

Differentiating both sides with respect to x, we get 

-e^-y\(\frac{dy}{dx}\) - e^-x = 0

\(\frac{dy}{dx}\) = \(\frac{e^{-x}}{-e^{-y}}\)

\(\frac{dy}{dx}\) = -ey - x

Concept:

Suppose that we have two functions f(x) and g(x) and they are both differentiable.

• Chain Rule: \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left[ {{\rm{f}}\left( {{\rm{g}}\left( {\rm{x}} \right)} \right)} \right] = {\rm{\;f'}}\left( {{\rm{g}}\left( {\rm{x}} \right)} \right){\rm{g'}}\left( {\rm{x}} \right)\)
• Product Rule: \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left[ {{\rm{f}}\left( {\rm{x}} \right){\rm{\;g}}\left( {\rm{x}} \right)} \right] = {\rm{\;f'}}\left( {\rm{x}} \right){\rm{\;g}}\left( {\rm{x}} \right) + {\rm{f}}\left( {\rm{x}} \right){\rm{\;g'}}\left( {\rm{x}} \right)\)
• log ab = b log a 

Calculation:-  

\(\rm x^m y^n =xya^{m-n}\), where a is constant 

Taking log both sides, we get

⇒ \(\rm \log (x^m y^n) =\log (xya^{m-n})\)

⇒ log xm + log yn = log x + log y + log am-n

⇒ m log x + n log y = log x + log y + (m - n) log a 

Differentiate both sides w.r.t  x, we get

⇒ \(\rm \frac{m}{x}+\frac{n}{y}\frac{\mathrm{d} y}{\mathrm{d} x}= \frac 1 x + \frac 1 y \frac {dy}{dx} + 0\)       [∵  \(\rm \frac{d \;\text{constant}}{dx} = 0\) ]  

⇒ \(\rm \frac{\mathrm{d} y}{\mathrm{d} x}\left [ \frac{n}{y}-\frac{1}{y} \right ]= \left [ \frac{1}{x}- \frac{m}{x} \right ]\) 

⇒  \(\rm \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{-(m-1)y}{(n-1)x}\) . 

The correct option is 1. 

Concept:

Quotient rule of differentiation:

\(\rm \frac {d \left(\frac {u}{v}\right) }{dx}= \frac {v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\)

Calculation:

We have xy = ex - y 

Taking log on both sides

⇒ y log x = (x - y) log e

⇒ y log x = x - y

⇒ (log x + 1)y = x

⇒ y = \(\rm x\over\log x+1\)

Differentiating with respect to x, we get

⇒ \(\rm {dy\over dx} = {(\log x+1)-x({1\over x})\over(\log x+1)^2}\)

⇒ \(\rm {dy\over dx} = {(\log x+1-1)\over(\log x+1)^2}\)

⇒ \(\boldsymbol{\rm {dy\over dx} = {\log x\over(\log x+1)^2}}\)

∴ The correct answer is \(\boldsymbol{\rm {dy\over dx} = {\log x\over(\log x+1)^2}}\).