Differentiation of Implicit Functions MCQ Quiz - Objective Question with Answer for Differentiation of Implicit Functions - Download Free PDF
Last updated on Jul 1, 2025
Latest Differentiation of Implicit Functions MCQ Objective Questions
Differentiation of Implicit Functions Question 1:
Comprehension:
Consider the following for the two (02) items that follow:
Let , where p,q are positive integers.
If , then what is \(\frac{dy}{dx}\) equal to?
Answer (Detailed Solution Below)
Differentiation of Implicit Functions Question 1 Detailed Solution
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.
Differentiation of Implicit Functions Question 2:
Comprehension:
Consider the following for the two (02) items that follow:
Let , where p,q are positive integers.
The derivative of y with respect to x
Answer (Detailed Solution Below)
Differentiation of Implicit Functions Question 2 Detailed Solution
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.
Differentiation of Implicit Functions Question 3:
Let (x + y)p + q = xpyp, where p, q are positive integers. If , then what is \(\frac{dy}{dx}\) equal to?
Answer (Detailed Solution Below)
Differentiation of Implicit Functions Question 3 Detailed Solution
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.
Differentiation of Implicit Functions Question 4:
Consider the following for the two (02) items that follow:
Let , where p,q are positive integers.
The derivative of y with respect to x
Answer (Detailed Solution Below)
Differentiation of Implicit Functions Question 4 Detailed Solution
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.
Differentiation of Implicit Functions Question 5:
If 2x + 2y = 2x+y, then \(\frac{d y}{d x}=\)
Answer (Detailed Solution Below)
Differentiation of Implicit Functions Question 5 Detailed Solution
Concept:
- \(\rm \frac{d}{dx}a^x=a^x\log a\)
Calculation:
Given 2x + 2y = 2x+y
We know that 2a+b = 2a⋅ 2b
⇒ 2x + 2y = 2x ⋅ 2y
⇒ \(\rm \frac{2^x+2^y}{2^x\cdot2^y}=1\)
⇒ 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
⇒ \(\rm \frac{dy}{dx}=-\frac{2^{-x}}{2^{-y}}\)
⇒ \(\rm \frac{dy}{dx}=-\frac{1-2^{-y}}{2^{-y}}\)
⇒ \(\rm \frac{dy}{dx}\) = - 2y + 1
⇒ \(\rm \frac{dy}{dx}\) = 1 - 2y
The required value of \(\rm \frac{dy}{dx}\) is 1 - 2y .
Top Differentiation of Implicit Functions MCQ Objective Questions
If xe = e\((\rm {x^2+y^2})\), then find \(\rm dy\over dx\)
Answer (Detailed Solution Below)
Differentiation of Implicit Functions Question 6 Detailed Solution
Download Solution PDFCalculation:
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(\(\rm 1\over x\)) = \(\rm 2x + 2y{dy\over dx}\)
⇒ \(\rm {e\over x}- 2x = 2y{dy\over dx}\)
∴ \(\rm {dy\over dx}={e- 2x^2\over2xy}\)
If 3x + 3y = 3x + y, then find \(\rm dy\over dx\).
Answer (Detailed Solution Below)
Differentiation of Implicit Functions Question 7 Detailed Solution
Download Solution PDFConcept:
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 \(\rm dy\over dx\)
⇒ \(\rm {dy\over dx}={3^{x+y}\ -\ 3^x\over3^y\ -\ 3^{x+y}}\)
If \(\rm 2x^3-3y^2=7\), what is \(\rm \dfrac{dy}{dx}\) equal to \(\rm (y ≠ 0)\)?
Answer (Detailed Solution Below)
Differentiation of Implicit Functions Question 8 Detailed Solution
Download Solution PDFCalculation:
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.
If y + sin-1 (1 - x2) = ex, then \(\rm {dy\over dx}\)
Answer (Detailed Solution Below)
Differentiation of Implicit Functions Question 9 Detailed Solution
Download Solution PDFCalculation:
y + sin-1 (1 - x2) = ex
y = ex - sin-1 (1 - x2)
Differentiating w.r.t x, we get
\(\rm {dy\over dx} = e^x - {1\over\sqrt{1-(1-x^2)^2}}(-2x)\)
\(\rm {dy\over dx} = e^x + {2x\over\sqrt{1-(1-2x^2+x^4)}}\)
\(\rm {dy\over dx} = e^x + {2x\over\sqrt{2x^2-x^4}}\)
\(\boldsymbol{\rm {dy\over dx} = e^x + {2\over\sqrt{2-x^2}}}\)
If \(\rm \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\)then \(\rm \frac{dy}{dx} = \)
Answer (Detailed Solution Below)
Differentiation of Implicit Functions Question 10 Detailed Solution
Download Solution PDFConcept:
\(\rm \frac{dx^n}{dx}=nx^{n-1}\)
Calculation:
Given: \(\rm \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\)
Differentiating with respect to x, we get
\(\rm \Rightarrow \frac{2x}{a^2} + \frac{2y}{b^2}\frac{dy}{dx} = 0\)
\(\rm \Rightarrow \frac{2y}{b^2}\frac{dy}{dx} = -\frac{2x}{a^2} \)
\(\rm \therefore \frac{dy}{dx} = -\frac{b^2x}{a^2y} \)
If y2 = 4ax then \(\rm \frac{dy}{dx} = \)
Answer (Detailed Solution Below)
Differentiation of Implicit Functions Question 11 Detailed Solution
Download Solution PDFConcept:
\(\rm \frac{dx^n}{dx}=nx^{n-1}\)
Calculation:
Given: y2 = 4ax
Differentiating with respect to x, we get
\(\rm \Rightarrow {2y}\frac{dy}{dx} = 4a\)
\(\rm \Rightarrow \frac{dy}{dx} = \frac{4a}{{2y}} \)
\(\rm \therefore \frac{dy}{dx} = \frac{2a}{y} \)
If y = 3e2x + 2e3x, then \(\frac{{{d^2}y}}{{d{x^2}}}\) - 5 \(\frac{{dy}}{{dx}}\) + 6y equals
Answer (Detailed Solution Below)
Differentiation of Implicit Functions Question 12 Detailed Solution
Download Solution PDFGiven
y = 3e2x + 2e3x
Formula used
d(xn)/dx = nxn-1
Solution
⇒dy/dx = 3e2x(2) + 2e3x(3)
⇒dy/dx = 6(e2x + e3x)
⇒d2y/dx2 = 6(2e2x+3e3x)
As asked in the question,
⇒ \(\frac{{{d^2}y}}{{d{x^2}}}\) - 5 \(\frac{{dy}}{{dx}}\) + 6y =
⇒ 12e2x + 18e3x − 30e2x − 30e3x + 18e2x + 12e3x
⇒ 0.
The correct option is 4.
What is the value of \(\rm dy\over dx\) , if y2 + x2 + 3x + 5 = 0 at (0, -3)?
Answer (Detailed Solution Below)
Differentiation of Implicit Functions Question 13 Detailed Solution
Download Solution PDFConcept:
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
The second derivative of the function y = f(x) given by the equation y2 = 2x is
Answer (Detailed Solution Below)
Differentiation of Implicit Functions Question 14 Detailed Solution
Download Solution PDFCalculation:
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)
\(\Rightarrow y'' = - \frac{{{{\left( {y'} \right)}^2}}}{y} = - \frac{1}{{{y^3}}}\)
If y = cos2 x2, find \(\frac {dy}{dx}\)
Answer (Detailed Solution Below)
Differentiation of Implicit Functions Question 15 Detailed Solution
Download Solution PDFConcept:
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.