Differential Calculus MCQ Quiz - Objective Question with Answer for Differential Calculus - Download Free PDF

Concept:

• \(\text{Related Rates}: \)  Problems where two or more quantities change with respect to time.
• \(\text{Volume of Cone } \) The formula is \(V = \frac{1}{3}\pi r^2 h\), where \(V\) is volume, \(r\) is radius, and \(h\) is height.
• \(\text{Constant Ratio Condition} : \) Given \(h = \frac{1}{6}r\), the radius and height are proportional, so express \(r\) in terms of \(h\).
• Differentiation: Differentiate \(V\) with respect to time \(t\) to find how fast the height changes as the volume increases.

Calculation:

Given,

\(\frac{dV}{dt} = 12 \, \text{cm}^3/\text{s}\)

Height and radius relation: \(h = \frac{1}{6}r\) so \(r = 6h\)

Volume of cone:

\(V = \frac{1}{3}\pi r^2 h\)

⇒ Substitute \(r = 6h\):

\(V = \frac{1}{3}\pi (6h)^2 h\)

⇒ \(V = \frac{1}{3}\pi \times 36h^2 \times h\)

⇒ \(V = 12\pi h^3\)

Differentiate both sides w.r.t \(t\):

\(\frac{dV}{dt} = 36\pi h^2 \frac{dh}{dt}\)

Substitute \(\frac{dV}{dt} = 12\) and \(h = 4\):

⇒ \(12 = 36\pi \times (4)^2 \times \frac{dh}{dt}\)

⇒ \(12 = 36\pi \times 16 \times \frac{dh}{dt}\)

⇒ \(12 = 576\pi \frac{dh}{dt}\)

⇒ \(\frac{dh}{dt} = \frac{12}{576\pi} = \frac{1}{48\pi}\)

The height increases at a rate of \(\frac{1}{48\pi} \, \text{cm/s}\). 

⇒ 96πh = 2

Hence 2 is the correct answer. 

Calculation:

Given,

\( (x + y)^{p+q} = x^p\,y^q \) and \( p + q = 10 \).

Differentiate both sides with respect to \(x\) implicitly:

\(\frac{d}{dx}\bigl((x+y)^{p+q}\bigr) = \frac{d}{dx}\bigl(x^p y^q\bigr) \)

Left side:

\((p+q)\,(x+y)^{p+q-1}\bigl(1 + \tfrac{dy}{dx}\bigr) \)

Right side (product rule):

\(p\,x^{p-1}y^q \;+\; q\,x^p y^{q-1}\,\tfrac{dy}{dx} \)

Rearrange to collect \( \tfrac{dy}{dx} \) terms and use

\( (x+y)^{p+q} = x^p y^q \implies (x+y)^{p+q-1} = \tfrac{x^p y^q}{x+y} \).

After cancellation of the common factor \( p\,y - q\,x \), you obtain:

∴ \( \frac{dy}{dx} = \frac{y}{x} \)

Hence, the correct answer is Option 1. 

Calculation:

Given,

\((x+y)^{p+q} = x^p\,y^q\)

Differentiate implicitly w.r.t. \(x\):

\((p+q)(x+y)^{p+q-1}\bigl(1+\tfrac{dy}{dx}\bigr) = p\,x^{p-1}y^q \;+\; q\,x^p\,y^{q-1}\tfrac{dy}{dx}\)

Rearrange to collect \(\tfrac{dy}{dx}\):

\(\tfrac{dy}{dx}\bigl[(p+q)(x+y)^{p+q-1} - q\,x^p\,y^{q-1}\bigr] = p\,x^{p-1}y^q - (p+q)(x+y)^{p+q-1}\)

Use \((x+y)^{p+q-1}=\frac{x^p\,y^q}{x+y}\) to simplify:

\(\tfrac{dy}{dx} = \tfrac{y}{x}\)

∴ \(\displaystyle \frac{dy}{dx} = \frac{y}{x}\), independent of \(p\) and \(q\).

Hence, the correct answer is Option 4.

Given:  f(x) = (x - 4) (x - 5)

Expanding the given equation 

⇒  f(x) = x² - 9x + 20

Differentiating both sides, we have:

f'(x) = 2x -9

Substituting the value of x = 5, we have

f'(5) = 2 x 5 - 9 = 1

Concept:

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

• Product Rule: \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left[ {{\rm{f}}\left( {\rm{x}} \right){\rm{\;g}}\left( {\rm{x}} \right)} \right] = {\rm{\;f'}}\left( {\rm{x}} \right){\rm{\;g}}\left( {\rm{x}} \right) + {\rm{f}}\left( {\rm{x}} \right){\rm{\;g'}}\left( {\rm{x}} \right)\)
• Division rule: \(\frac{d}{dx}\frac{u}{v}=\frac{v\times u'-u \times v'}{v^2}\)

Formulas:

\(\rm \frac{d \cot^{-1} x}{dx} = \frac{-1}{1+x^2}\)

Calculation:

\(\rm \frac {d^2 \cot^ {-1}x}{dx^2}\)

= \(\rm \frac{d}{dx} \times \frac {d\cot^{-1} x}{dx}\)

= \(\frac{d}{dx}(\frac{-1}{1+x^2})\)

= \(-\:\frac{d}{dx}(\frac{1}{1+x^2})\)

⇒ \(-[\frac{(1+x^2)\times 0-1 \times2x}{(1+x^2)^2}]\)

⇒\(-[\rm \frac{-2x}{(1+x^2)^2}] \)

= \(\rm \frac{2x}{(1+x^2)^2} \)

Concept:

The equation of the tangent to a curve y = f(x) at a point (a, b) is given by (y - b) = m(x - a), where m = y'(b) = f'(a) [value of the derivative at point (a, b)].

Calculation:

⇒ y' = f'(x) = 3x²

m = f'(1) = 3 × 1² = 3.

(y - b) = m(x - a)

⇒ (y - 1) = 3(x - 1)

⇒ y - 1 = 3x - 3

Calculations:

Given, \(\rm f(x) = x- \dfrac{1}{x}\)

Differentiating with respect to x, we get

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

⇒ 1 + \(\rm\frac {1}{x^2}\)

Put x = -1

⇒ f'(-1) = 1 + \(\rm\frac {1}{(-1)^2}\) = 1 + 1 = 2

∴ f'(-1) = 2

Concept:

Following steps to finding minima using derivatives.

• Find the derivative of the function.
• Set the derivative equal to 0 and solve. This gives the values of the maximum and minimum points.
• Now we have to find the second derivative: If f"(x) Is greater than 0 then the function is said to be minima

Calculation:

f(x) = x2 - x + 2

f'(x) = 2x - 1

Set the derivative equal to 0, we get

f'(x) = 2x - 1 = 0

⇒ x = \(\frac12\)

Now, f''(x) = 2 > 0

So, we get minimum value at x = \(\frac12\)

f(\(\frac12\)) = (\(\frac12\))² - \(\frac12\) + 2 = \(\frac74\)

Hence, option (3) is correct. 

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: \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left[ {{\rm{f}}\left( {\rm{x}} \right){\rm{\;g}}\left( {\rm{x}} \right)} \right] = {\rm{\;f'}}\left( {\rm{x}} \right){\rm{\;g}}\left( {\rm{x}} \right) + {\rm{f}}\left( {\rm{x}} \right){\rm{\;g'}}\left( {\rm{x}} \right)\)

Calculation:

y = x^x

Taking log both sides, we get

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

⇒ log y = x log x

Differentiating with respect to x, we get

\(\rm \Rightarrow \frac{1}{y} \times \frac{dy}{dx} = x \times \frac{\log x}{dx} + \log x \times \frac{dx}{dx}\)

\(\rm \Rightarrow \frac{1}{y} \times \frac{dy}{dx} = x \times \frac{1}{x} + \log x \times1\)

\(\rm \Rightarrow \frac{dy}{dx} = y \times [1 + \log x]\)

\(\rm \Rightarrow \frac{dy}{dx} = x^x \times [1 + \log x]\)

put x = 1

\(\rm \Rightarrow \frac{dy}{dx} = 1^1 \times [1 + \log 1]\)

\(\rm \Rightarrow \frac{dy}{dx} = 1 \times [1 + 0]\)                (∵ log 1 = 0)

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

Concept:

Rate of change of 'x' is given by \(\rm \frac {dx}{dt}\)

Calculation:

Given that, y = 2x – x² and \(\rm \frac {dx}{dt}\) = 3 units/sec

Then, the slope of the curve, \(\rm \frac {dy}{dx}\) = 2 - 2x = m

⇒\(\rm \frac {dm}{dt}\)  = 0 - 2 × \(\rm \frac {dx}{dt}\)

= -2(3)

= -6 units per second

Hence, the slope of the curve is decreasing at the rate of 6 units per second when x is increasing at the rate of 3 units per second.

Hence, option (2) is correct.

Concept:

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

• Chain Rule: \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left[ {{\rm{f}}\left( {{\rm{g}}\left( {\rm{x}} \right)} \right)} \right] = {\rm{\;f'}}\left( {{\rm{g}}\left( {\rm{x}} \right)} \right){\rm{g'}}\left( {\rm{x}} \right)\)

Calculation:

Given:

y = sin x°

We know that,

180° = π radian

∴ 1° = \(\frac{\pi}{180}\) radian

Now, x° = \(\frac{\pi x}{180}\) radian

⇒ y = \(\sin \left( {\frac{{\rm{\pi \cdot x }}}{{180}}} \right){\rm{\;}}\)

Differenatiate with respect to x, we get

\(\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = {\rm{\;}}\frac{{\rm{\pi }}}{{180}}{\rm{\;}} \times \cos \left( {\frac{{\rm{\pi \cdot x}}}{{180}}} \right){\rm{\;}}\)

Calculation:

Given: x = t² , y = t³.

⇒ \(\frac{dx}{dt} =2t\) and \(\frac{dy}{dt}=3t^2\)

\(\frac{dy}{dx}=\frac{dy/dt}{dx/dt}\)

\(\frac{dy}{dx}=\frac{3t^2}{2t} =\frac{3}{2}t\)

Again differentiating with respect to x:

⇒ \(\frac{d^2y}{dx^2}=\frac{3}{2}\cdot\frac{dt}{dx}\)

⇒ \(\frac{d^2y}{dx^2}=\frac{3}{2}\cdot\frac{1}{2t}\)  (∵ \(\frac{dx}{dt} =2t\))

∴  \(\frac{d^2y}{dx^2}=\frac{3}{4t}\)

The correct answer is \(\frac{3}{4t}\).

Concept:

Chain Rule of Derivatives: 

\(\rm \dfrac{d}{dx}f(g(x))=\dfrac{d}{d\ g(x)}f(g(x))\times \dfrac{d}{dx}g(x)\).

\(\rm \dfrac{d}{dx}e^x\) = e^x.

Calculation:

It is given that y = \(\rm e^{x+e^{x+e^{x+^{\ ...\ \infty}}}}\).

∴ y = \(\rm e^{x+(e^{x+e^{x+^{\ ...\ \infty}}})}=e^{x+y}\)

Differentiating both sides with respect to x and using the chain rule, we get:

\(\rm \dfrac{dy}{dx}=\dfrac{d}{dx}e^{x+y}\)

⇒ \(\rm \dfrac{dy}{dx}=e^{x+y}\dfrac{d}{dx}(x+y)\)

⇒ \(\rm \dfrac{dy}{dx}=y\left (1+\dfrac{dy}{dx} \right )\)

⇒ \(\rm \dfrac{dy}{dx}=y+y\dfrac{dy}{dx}\)

⇒ \(\rm (1-y)\dfrac{dy}{dx}=y\)

⇒ \(\rm \dfrac{dy}{dx}=\dfrac{y}{1-y}\).

Concept:

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

Calculation:

Given:  y = e^{log (log x)}

To Find: \(\rm \frac{dy}{dx}\)

As we know that, e^{log x} = x

∴ e^{log (log x)} = log x

Now, y = log x

Differentiating with respect to x, we get

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