Partial Differentiation Questions and Answers – Variable Treated as Constant

«
»

This set of Differential and Integral Calculus Multiple Choice Questions & Answers (MCQs) focuses on “Variable Treated as Constant”.

1. If z=3xy+4x2, what is the value of \(\frac{∂z}{∂x}\)?
a) 3y+8x
b) 3x+4x2
c) 3xy+8x
d) 3y+3x+8x
View Answer

Answer: a
Explanation: Given: z=3xy+4x2
Using partial differentiation, we need to differentiate the function z with respect to x keeping y as constant. Thus, \(\frac{∂z}{∂x}=3y+8x.\)
advertisement

2. The value of \(\frac{∂z}{∂y}\)=8x2+6xy2+4. What is the function z expressed as?
a) z=8x3+2x2 y2+4x
b) z=8x2 y+2xy3+4y
c) z=8y+2xy2+4y
d) z=16x+6y2
View Answer

Answer: b
Explanation: Since the given partial differentiation is with respect to y, to find the expression for the function z, we need to integrate the given partial differentiation expression with respect to y.
Therefore, \(∫\frac{∂z}{∂y}.dy\)=∫(8x2+6xy2+4).dy
z=8x2 y+\(\frac{6}{3}\) xy3+4y
z=8x2 y+2xy3+4y

3. Find the correct values for \(\frac{∂f}{∂x} \,and\, \frac{∂f}{∂y}\) for the function \(f=\frac{2}{x^3}y^2+4y^3.\)
a) \(\frac{∂f}{∂x}= \frac{-6}{x^2}, \frac{∂f}{∂y}= \frac{2}{x^3} y+8y^2\)
b) \(\frac{∂f}{∂x}= \frac{2}{x^4}, \frac{∂f}{∂y}= \frac{2}{x^3} y+12y^2\)
c) \(\frac{∂f}{∂x}= \frac{-6}{x^4}, \frac{∂f}{∂y}= \frac{4}{x^3} y+12y^2\)
d) \(\frac{∂f}{∂x}= \frac{-6}{x^4}, \frac{∂f}{∂y}= \frac{4}{x^3} y^2+12\)
View Answer

Answer: c
Explanation: Given: \(f=\frac{2}{x^3}y^2+4y^3\)
Differentiating partially with respect to x we get,
\(\frac{∂f}{∂x}=2\frac{∂(x^{-3})}{∂x}=2(-3x^{-3-1}) = \frac{-6}{x^4}\)
Differentiating with respect to y we get,
\(\frac{∂f}{∂y}= \frac{4}{x^3}y+12y^2\)
Note: Join free Sanfoundry classes at Telegram or Youtube
advertisement
advertisement

4. Volume of an object expressed in spherical coordinates is given by \(V = \int_0^{\frac{π}{2}}∫_0^{\frac{π}{3}}∫_0^1r cos∅ dr d∅ dθ.\) The value of the integral is _______
a) \(\frac{\sqrt{3}}{2}\)
b) \(\frac{1}{√2} π\)
c) \(\frac{\sqrt{3}}{2} π\)
d) \(\frac{\sqrt{3}}{8}π\)
View Answer

Answer: d
Explanation: Given: \(V = \int_0^{\frac{π}{2}}∫_0^{\frac{π}{3}}∫_0^1r cos∅ \,dr \,d∅ \,dθ\)
\(V = \int_0^{\frac{π}{2}}∫_0^{\frac{π}{3}}(\frac{r^2}{2})_0^1cos∅ \,d∅ \,dθ\)
\(V = \frac{1}{2} \int_0^{\frac{π}{2}}(sin∅)_0^{\frac{π}{3}}d∅ \,dθ\)
\(V = \frac{1}{2}×\frac{\sqrt{3}}{2}\int_0^{\frac{π}{2}}dθ\)
\(V = \frac{1}{2}×\frac{\sqrt{3}}{2}×\frac{π}{2}\)
\(V = \frac{\sqrt{3}}{8}π \)

5. Which of the following relations hold true for division rule of differentiation?
a) \((\frac{f(x)}{g(x)})’= \frac{f'(x)}{g'(x)} \)
b) \((\frac{f(x)}{g(x)})’= \frac{g(x) f'(x)- g'(x)f(x)}{(f(x))^2}\)
c) \((\frac{f(x)}{g(x)})’= \frac{g(x)f'(x)- g'(x)f(x)}{(g(x))^2} \)
d) \((\frac{f(x)}{g(x)})’= \frac{f(x)g'(x)-f'(x)g(x)}{(g(x))^2} \)
View Answer

Answer: c
Explanation: The division rule of differentiation for two functions is given by,
\((\frac{f(x)}{g(x)})’= \frac{g(x)f'(x)- g'(x)f(x)}{(g(x))^2} \)
advertisement

6. What is the value of \(\frac{∂^2z}{∂x∂y}\) for the z=3x2y+5y?
a) 3xy
b) 6x
c) 3x+5
d) 6xy
View Answer

Answer: b
Explanation: Given: z=3x2y+5y
\(\frac{∂}{∂x}(\frac{∂z}{∂y})=\frac{∂}{∂x}(3x^2+5)=6x \)

7. Which of the following is correct?
a) \(\frac{d}{dx} (sin^{-1}(⁡x))= \frac{1}{\sqrt{1-x^2}}\)
b) \(\frac{d}{dx} (sec^{-1}(⁡x))= \frac{1}{\sqrt{x^2-1}}\)
c) \(\frac{d}{dx} (tan^{-1}(⁡x))= \frac{1}{\sqrt{x^2+1}}\)
d) \(\frac{d}{dx} (sin^{-1}(⁡x))= \frac{1}{x+1} \)
View Answer

Answer: a
Explanation: Rules for derivatives of inverse trigonometric functions are:

  • \(\frac{d}{dx} (sin^{-1}⁡(⁡x))= \frac{1}{\sqrt{1-x^2}}\)
  • \(\frac{d}{dx} (sec^{-1}⁡(⁡x))= \frac{1}{x\sqrt{x^2-1}}\)
  • \(\frac{d}{dx} (tan^{-1}(⁡x))= \frac{1}{1+x^2}\)
advertisement

8. Given \(∫_0^8x^{\frac{1}{3}}dx,\) find the error in approximating the integral using Simpson’s 1/3 Rule with n=4.
a) 1.8
b) 2.9
c) 0.3
d) 0.35
View Answer

Answer: d
Explanation: Given: \(∫_0^8x^{\frac{1}{3}}dx, n = 4,\)
Let \(f(x)= x^{\frac{1}{3}},\)
\(∆x = \frac{b-a}{2}=\frac{8-0}{2}=4\) ………………since b=8, a=0 (limits of the given integral)
Hence endpoints xi have coordinates {0, 2, 4, 6, 8}.
Calculating the function values at xi, we get,
\(f(0)= 0^{\frac{1}{3}}=0\)
\(f(2)= 2^{\frac{1}{3}}\)
\(f(4)= 4^{\frac{1}{3}}\)
\(f(6)= 6^{\frac{1}{3}}\)
\(f(8)= 8^{\frac{1}{3}} =2\)
Substituting these values in the formula,
\(∫_0^8x^{\frac{1}{3}}dx ≈ \frac{∆x}{3} [f(0)+4f(2)+2f(4)+4f(6)+f(8)]\)
\( ≈ \frac{2}{3} [0+4(2^{\frac{1}{3}})+2(4^{\frac{1}{3}})+ 4(6^{\frac{1}{3}})+2] ≈ 11.65\)
Actual integral value,
\( ∫_0^8x^{\frac{1}{3}}dx= \left(\frac{x^{\frac{4}{3}}}{\frac{4}{3}}\right)_0^8=12\)
Error in approximating the integral = 12 – 11.65 = 0.35

9. The determinant of the matrix whose eigen values are 6, 4, 3 is given by ___________
a) 3
b) 24
c) 72
d) 13
View Answer

Answer: b
Explanation: The product of the eigen values of a matrix gives the determinant of the matrix,
Therefore, ∆ = 72.
advertisement

10. The symbol used for partial derivatives, ∂, was first used in mathematics by Marquis de Condorcet.
a) True
b) False
View Answer

Answer: a
Explanation: Partial derivatives are indicated by the symbol ∂. This was first used in mathematics by Marquis de Condorcet who used it for partial differences.

Sanfoundry Global Education & Learning Series – Differential and Integral Calculus.

To practice all areas of Differential and Integral Calculus, here is complete set of 1000+ Multiple Choice Questions and Answers.

advertisement
advertisement

Leave a Comment

Subscribe to our Newsletters (Subject-wise). Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

Youtube | Telegram | LinkedIn | Instagram | Facebook | Twitter | Pinterest
Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

Subscribe to his free Masterclasses at Youtube & technical discussions at Telegram SanfoundryClasses.