Differentiation of Implicit Functions MCQ Quiz - Objective Question with Answer for Differentiation of Implicit Functions - Download Free PDF

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.

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.

Concept:

• \(\rm \frac{d}{dx}a^x=a^x\log a\)

Calculation:

Given 2x + 2y = 2x+y

We know that 2^a+b = 2^a⋅ 2^b

⇒ 2^x + 2^y = 2^x ⋅ 2^y

⇒ \(\rm \frac{2^x+2^y}{2^x\cdot2^y}=1\)

⇒ 2^-y + 2^-x = 1  ..(1)

Differentiating the above equation with respect to x:

⇒ (-2^{- y}\(\rm\frac{dy}{dx}\) - 2^{- x}) log 2 = 0

⇒ \(\rm \frac{dy}{dx}=-\frac{2^{-x}}{2^{-y}}\)

⇒ \(\rm \frac{dy}{dx}=-\frac{1-2^{-y}}{2^{-y}}\)

⇒ \(\rm \frac{dy}{dx}\) = - 2^y + 1

⇒ \(\rm \frac{dy}{dx}\) = 1 - 2^y

The required value of  \(\rm \frac{dy}{dx}\) is 1 - 2^y .  

Calculation:

xe = e\((\rm x^2+y^2)\) 

Taking log on both sides, we get

⇒ log xe = log e\(\rm x^2+y^2\) 

⇒ e log x = (x² + y²) log e             

[∵ log mⁿ = n log m]

⇒ e log x = x2 + y2                       

[∵ log e = 1]

Differentiating w.r.t x, we get

⇒ e(\(\rm 1\over x\)) = \(\rm 2x + 2y{dy\over dx}\)

⇒ \(\rm {e\over x}- 2x = 2y{dy\over dx}\)

∴ \(\rm {dy\over dx}={e- 2x^2\over2xy}\)

Concept:

Calculus:

• \(\rm {d\over dx}\left(a^x\right)=a^x(\log a)\).
•  

Chain Rule of Derivatives:

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

Calculation:

Given expression is:

3x + 3y = 3x + y.

Differentiating w.r.t. x and using the chain rule of derivatives, we get:

⇒ 3x (log 3) + 3y (log 3) \(\rm dy\over dx\) = 3x + y (log 3) (1 + \(\rm dy\over dx\))

⇒ 3x + 3y \(\rm dy\over dx\) = 3x + y + 3x + y \(\rm dy\over dx\)

⇒ \(\rm {dy\over dx}={3^{x+y}\ -\ 3^x\over3^y\ -\ 3^{x+y}}\)

Calculation:

Here, 

\(\rm 2x^3-3y^2=7\)

Differentiating w.r.t. x, we get

\(\rm 6x^2-6y\frac{dy}{dx}=0\\ \Rightarrow \rm x^2-y\frac{dy}{dx}=0\\ \Rightarrow \frac{dy}{dx}=\rm \frac{x^2}{y}\\\)

Hence, option (3) is correct.

Calculation:

y + sin-1 (1 - x2) = ex 

y = e^x - sin^-1 (1 - x²)

Differentiating w.r.t x, we get

\(\rm {dy\over dx} = e^x - {1\over\sqrt{1-(1-x^2)^2}}(-2x)\)

\(\rm {dy\over dx} = e^x + {2x\over\sqrt{1-(1-2x^2+x^4)}}\)

\(\rm {dy\over dx} = e^x + {2x\over\sqrt{2x^2-x^4}}\)

\(\boldsymbol{\rm {dy\over dx} = e^x + {2\over\sqrt{2-x^2}}}\)

Concept:

\(\rm \frac{dx^n}{dx}=nx^{n-1}\)

Calculation:

Given: \(\rm \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\)

Differentiating with respect to x, we get

\(\rm \Rightarrow \frac{2x}{a^2} + \frac{2y}{b^2}\frac{dy}{dx} = 0\)

\(\rm \Rightarrow \frac{2y}{b^2}\frac{dy}{dx} = -\frac{2x}{a^2} \)

\(\rm \therefore \frac{dy}{dx} = -\frac{b^2x}{a^2y} \)

Concept:

\(\rm \frac{dx^n}{dx}=nx^{n-1}\)

Calculation:

Given: y2 = 4ax

Differentiating with respect to x, we get

\(\rm \Rightarrow {2y}\frac{dy}{dx} = 4a\)

\(\rm \Rightarrow \frac{dy}{dx} = \frac{4a}{{2y}} \)

\(\rm \therefore \frac{dy}{dx} = \frac{2a}{y} \)

Given

y = 3e^2x + 2e^3x

Formula used

d(xⁿ)/dx = nx^n-1

Solution

⇒dy/dx = 3e^2x(2) + 2e^3x(3)

⇒dy/dx = 6(e^2x + e^3x)

⇒d²y/dx² = 6(2e^2x+3e^3x)

As asked in the question,

⇒  \(\frac{{{d^2}y}}{{d{x^2}}}\) - 5 \(\frac{{dy}}{{dx}}\) + 6y =

⇒ 12e^2x + 18e^3x − 30e^2x − 30e^3x + 18e^2x + 12e^3x

⇒ 0.

The correct option is 4.

Concept:

Chain Rule (Differentiation by substitution): If y is a function of u and u is a function of x

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

Calculation:

Given y2 + x2 + 3x + 5 = 0

Differentiating with respect to x, we get

2y \(\rm dy\over dx\) + 2x +3(1) + 0 = 0

2y\(\rm dy\over dx\) + 2x + 3 = 0 

2y \(\rm dy\over dx\) = -(2x + 3)

\(\rm {dy\over dx} = -{2x+3\over2y}\)

Now at (0, -3)

\(\rm {dy\over dx} = -{2(0)+3\over2(-3)}\)

\(\rm {dy\over dx} = -{3\over(-6)}\)

\(\rm {dy\over dx} = {1\over2}\) = 0.5

Calculation:

Given function is y² = 2x;

Differentiating with respect to x on both sides,

2yy' = 2x ⇒ yy’ = 1   - (1)

⇒ y’ = 1/y;

Differentiating (1) again with respect to x on both sides,

⇒ y . y” + y’. y’ = 0   - (2)

\(\Rightarrow y'' = - \frac{{{{\left( {y'} \right)}^2}}}{y} = - \frac{1}{{{y^3}}}\)

Concept:

cos2x = 2cos²x - 1

sin2x = 2sin x cos x

Calculation:

Here, y = cos2 x2

Let, x² = t 

Differentiating with respect to x, we get

⇒2xdx = dt 

⇒ dt/dx = 2x ....(1)

y = cos²t

=\(\rm \frac{\cos2t+1}{2}=\frac{\cos2t}{2}+\frac{1}{2}\)

\(\rm \frac{dy}{dx} = \frac{1}{2}\frac{d}{dt}(\cos2t)\frac{dt}{dx}+0\\ = \frac{1}{2}(-2\sin2t)\frac{dt}{dx}\cdots (from \ (1))\)

= - sin2x² × 2x

= -4x cos x2 sin x2

Hence, option (2) is correct.