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, \(\frac{∂z}{∂x}=3y+8x.\)

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,
\(\frac{∂f}{∂y}= \frac{4}{x^3}y+12y^2\)

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,
\((\frac{f(x)}{g(x)})’= \frac{g(x)f'(x)- g'(x)f(x)}{(g(x))^2} \)
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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:

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

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.

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.

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