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:

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

Left side:

$(p+q)(x+y{)}^{p+q-1}(1+\frac{dy}{dx})$

Right side (product rule):

$p{x}^{p-1}{y}^q+q{x}^p{y}^{q-1}\frac{dy}{dx}$

Rearrange to collect $\frac{dy}{dx}$ terms and use

$(x+y{)}^{p+q}=x^p{y}^q⟹(x+y{)}^{p+q-1}=\frac{x^p{y}^q}{x+y}$.

After cancellation of the common factor $py-qx$, 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. $x$:

$(p+q)(x+y{)}^{p+q-1}(1+\frac{dy}{dx})=p{x}^{p-1}{y}^q+q{x}^p{y}^{q-1}\frac{dy}{dx}$

Rearrange to collect $\frac{dy}{dx}$:

$\frac{dy}{dx}[(p+q)(x+y{)}^{p+q-1}-q{x}^p{y}^{q-1}]=p{x}^{p-1}{y}^q-(p+q)(x+y{)}^{p+q-1}$

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

$\frac{dy}{dx}=\frac{y}{x}$

∴ ${\displaystyle \frac{dy}{dx}=\frac{y}{x}}$, 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: $\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\left[\mathrm{f}\left(\mathrm{x}\right)\mathrm{g}\left(\mathrm{x}\right)\right]={\mathrm{f}}^{\prime }\left(\mathrm{x}\right)\mathrm{g}\left(\mathrm{x}\right)+\mathrm{f}\left(\mathrm{x}\right){\mathrm{g}}^{\prime }\left(\mathrm{x}\right)$
• Division rule: $\frac{d}{dx}\frac{u}{v}=\frac{v\times u^{\prime }-u\times v^{\prime }}{v^2}$

Formulas:

$\frac{\mathrm{d}{\cot }^{-1}\mathrm{x}}{\mathrm{d}\mathrm{x}}=\frac{-1}{1+{\mathrm{x}}^2}$

Calculation:

$\frac{{\mathrm{d}}^2{\cot }^{-1}\mathrm{x}}{\mathrm{d}{\mathrm{x}}^2}$

= $\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\times \frac{\mathrm{d}{\cot }^{-1}\mathrm{x}}{\mathrm{d}\mathrm{x}}$

= $\frac{d}{dx}(\frac{-1}{1+x^2})$

= $-\frac{d}{dx}(\frac{1}{1+x^2})$

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

⇒$-[\frac{-2\mathrm{x}}{(1+{\mathrm{x}}^2{)}^2}]$

= $\frac{2\mathrm{x}}{(1+{\mathrm{x}}^2{)}^2}$

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, $\mathrm{f}(\mathrm{x})=\mathrm{x}-{\displaystyle \frac{1}{\mathrm{x}}}$

Differentiating with respect to x, we get

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

⇒ 1 + $\frac{1}{{\mathrm{x}}^2}$

Put x = -1

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

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

So, we get minimum value at x = $\frac{1}{2}$

f($\frac{1}{2}$) = ($\frac{1}{2}$)² - $\frac{1}{2}$ + 2 = $\frac{7}{4}$

Hence, option (3) is correct. 

Concept:

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

• Chain Rule: $\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\left[\mathrm{f}\left(\mathrm{g}\left(\mathrm{x}\right)\right)\right]={\mathrm{f}}^{\prime }\left(\mathrm{g}\left(\mathrm{x}\right)\right){\mathrm{g}}^{\prime }\left(\mathrm{x}\right)$
• Product Rule: $\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\left[\mathrm{f}\left(\mathrm{x}\right)\mathrm{g}\left(\mathrm{x}\right)\right]={\mathrm{f}}^{\prime }\left(\mathrm{x}\right)\mathrm{g}\left(\mathrm{x}\right)+\mathrm{f}\left(\mathrm{x}\right){\mathrm{g}}^{\prime }\left(\mathrm{x}\right)$

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

$\Rightarrow \frac{1}{\mathrm{y}}\times \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\mathrm{x}\times \frac{\log \mathrm{x}}{\mathrm{d}\mathrm{x}}+\log \mathrm{x}\times \frac{\mathrm{d}\mathrm{x}}{\mathrm{d}\mathrm{x}}$

$\Rightarrow \frac{1}{\mathrm{y}}\times \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\mathrm{x}\times \frac{1}{\mathrm{x}}+\log \mathrm{x}\times 1$

$\Rightarrow \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\mathrm{y}\times [1+\log \mathrm{x}]$

$\Rightarrow \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}={\mathrm{x}}^{\mathrm{x}}\times [1+\log \mathrm{x}]$

put x = 1

$\Rightarrow \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=1^1\times [1+\log 1]$

$\Rightarrow \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=1\times [1+0]$                (∵ log 1 = 0)

$\therefore \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=1$

Concept:

Rate of change of 'x' is given by $\frac{\mathrm{d}\mathrm{x}}{\mathrm{d}\mathrm{t}}$

Calculation:

Given that, y = 2x – x² and $\frac{\mathrm{d}\mathrm{x}}{\mathrm{d}\mathrm{t}}$ = 3 units/sec

Then, the slope of the curve, $\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}$ = 2 - 2x = m

⇒$\frac{\mathrm{d}\mathrm{m}}{\mathrm{d}\mathrm{t}}$  = 0 - 2 × $\frac{\mathrm{d}\mathrm{x}}{\mathrm{d}\mathrm{t}}$

= -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: $\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\left[\mathrm{f}\left(\mathrm{g}\left(\mathrm{x}\right)\right)\right]={\mathrm{f}}^{\prime }\left(\mathrm{g}\left(\mathrm{x}\right)\right){\mathrm{g}}^{\prime }\left(\mathrm{x}\right)$

Calculation:

Given:

y = sin x°

We know that,

180° = π radian

∴ 1° = $\frac{\pi }{180}$ radian

Now, x° = $\frac{\pi x}{180}$ radian

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

Differenatiate with respect to x, we get

$\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\frac{\pi }{180}\times \cos \left(\frac{\pi \cdot \mathrm{x}}{180}\right)$

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:

⇒ $\frac{d^{2}y}{d{x}^2}=\frac{3}{2}\cdot \frac{dt}{dx}$

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

∴  $\frac{d^{2}y}{d{x}^2}=\frac{3}{4t}$

The correct answer is $\frac{3}{4t}$.

Concept:

Chain Rule of Derivatives: 

${\displaystyle \frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}}\mathrm{f}(\mathrm{g}(\mathrm{x}))={\displaystyle \frac{\mathrm{d}}{\mathrm{d}\mathrm{g}(\mathrm{x})}}\mathrm{f}(\mathrm{g}(\mathrm{x}))\times {\displaystyle \frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}}\mathrm{g}(\mathrm{x})$.

${\displaystyle \frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}}{\mathrm{e}}^{\mathrm{x}}$ = e^x.

Calculation:

It is given that y = ${\mathrm{e}}^{\mathrm{x}+{\mathrm{e}}^{\mathrm{x}+{\mathrm{e}}^{\mathrm{x}{+}^{...\mathrm{\infty }}}}}$.

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

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

${\displaystyle \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}}={\displaystyle \frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}}{\mathrm{e}}^{\mathrm{x}+\mathrm{y}}$

⇒ ${\displaystyle \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}}={\mathrm{e}}^{\mathrm{x}+\mathrm{y}}{\displaystyle \frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}}(\mathrm{x}+\mathrm{y})$

⇒ ${\displaystyle \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}}=\mathrm{y}(1+{\displaystyle \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}})$

⇒ ${\displaystyle \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}}=\mathrm{y}+\mathrm{y}{\displaystyle \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}}$

⇒ $(1-\mathrm{y}){\displaystyle \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}}=\mathrm{y}$

⇒ ${\displaystyle \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}}={\displaystyle \frac{\mathrm{y}}{1-\mathrm{y}}}$.

Concept:

$\frac{\mathrm{d}(\log \mathrm{x})}{\mathrm{d}\mathrm{x}}=\frac{1}{\mathrm{x}}$

Calculation:

Given:  y = e^{log (log x)}

To Find: $\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}$

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

$\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\frac{\mathrm{d}(\log \mathrm{x})}{\mathrm{d}\mathrm{x}}=\frac{1}{\mathrm{x}}$