Partial Differentiation Questions and Answers – Total Derivative

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This set of Differential and Integral Calculus Multiple Choice Questions & Answers (MCQs) focuses on “Total Derivative”.

1. The total derivative is the same as the derivative of the function.
a) True
b) False
View Answer

Answer: a
Explanation: In mathematics, the total derivative of a function at a point is the best linear approximation near this point of the function with respect to its arguments.
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2. What is the derivative of \(7\sqrt[3]{x}-\frac{3}{x^4}+5x\) with respect x?
a) \(7x^{\frac{-1}{3}}-3x^{-5}+5\)
b) \(7x^{\frac{-2}{3}}-3x^{-5}+5\)
c) \(7x^{\frac{-2}{3}}-3x^{-3}+5\)
d) \(7x^{\frac{-1}{3}}-3x^{-3}+5\)
View Answer

Answer: b
Explanation: Given: y= \(7\sqrt[3]{x}-\frac{3}{x^4}+5x\)
\(\frac{dy}{dx}=\frac{d(7x^{\frac{-1}{3}}-3x^{-4}+5x)}{dx}\)
\(\frac{dy}{dx}=7x^{\frac{-1}{3}-1}-3x^{-4-1}+5x^{1-1}\)
\(\frac{dy}{dx}=7x^{\frac{-2}{3}}-3x^{-5}+5\)

3. Find the range in which the function f(x) = 8 + 40x3 – 5x4 – 4x5 is increasing.
a) 2<z<0, 0<z<3
b) 1<z<0, 0<z<2
c) 3<z<0, 0<z<2
d) 3<z<0, 0<z<4
View Answer

Answer: c
Explanation: Given: f(x)=8 + 40x3 – 5x4 – 4x5
f'(x) = 120x2 – 20x3 – 20x4
f'(x) = -20x2 (x+x2 – 6)
f'(x) = -20x2 (x+3)(x-2)
Next, we need to know where the function is not changing and so all we need to do is set the derivative equal to zero and solve.
f'(x) = -20x2 (x+3)(x-2)=0
From this it is pretty easy to see that the derivative will be zero, and hence the function will not be moving, at,
x=0,-3,-2
Because the derivative is continuous, we know that the only place it can change sign is where the derivative is zero. So, a quick number line will give us the sign of the derivative for the various intervals.

From this we get,
Increasing: 3<z<0, 0<z<2

4. What is the maximum area of the rectangle with perimeter 500 mm?
a) 15,625 mm2
b) 15,025 mm2
c) 15,600 mm2
d) 10,625 mm2
View Answer

Answer: a
Explanation: Let x be the length of the rectangle and y be the width of the rectangle. Then, Area A is,
A=x*y …………………………………………………. (1)
Given: Perimeter of the rectangle is 620 mm. Therefore,
P=2(x+y)
500=2(x+y)
x+y=250
y=250-x
We can now substitute the value of y in (1)
A=x*(250-x)
A=250x-x2
To find maximum value we need derivative of A,
\(\frac{dA}{dx}=250-2x\)
To find maximum value, \(\frac{dA}{dx}=0\)
250-2x=0
2x=250
x=125 mm
Therefore, when the value of x=125 mm and the value of y=250-125=125 mm, the area of the rectangle is maximum, i.e., A=125*125=15,625 mm2

5. Which of the following relations hold true?
a) i × i = j × j = k × k = 1
b) i × j = k, j × i = -k
c) i × i = j × j = k × k = -1
d) k × i = -j, i × k = j
View Answer

Answer: b
Explanation: The properties of vector or cross product, for the orthogonal vectors, i, j, and k are,
i × i = j × j = k × k = 0,
i × j = k, j × i = -k,
j × k = i, k × j = -i,
k × i = j, i × k = -j
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6. Which of the following trigonometric function derivatives is correct?
a) \(\frac{d(sinx)}{dx}=-cosx\)
b) \(\frac{d(secx)}{dx}=tanx\)
c) \(\frac{d(tanx)}{dx}=sec^2 x\)
d) \(\frac{d(cosx)}{dx}=sinx\)
View Answer

Answer: c
Explanation: Correct forms of Trigonometric Derivative Functions

  • \(\frac{d(sinx)}{dx}=cosx\)
  • \(\frac{d(cosx)}{dx}=-sinx\)
  • \(\frac{d(secx)}{dx}=secxtanx\)
  • \(\frac{d(tanx)}{dx}=sec^2 x\)

7. The division rule of differentiation for two functions is given by, \((\frac{f(x)}{g(x)})’= \frac{f'(x)-g’ (x)}{(g(x))^2}. \)
a) True
b) False
View Answer

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

8. What is the derivative of z=3x*logx+5x6 ex with respect to x?
a) 3+30x5 ex
b) 3+5x6 ex+30x5 ex
c) 3+5x6 ex
d) 3+3logx+5x6 ex+30x5 ex
View Answer

Answer: d
Explanation: Given: z=3x*logx+5x6 ex
\(\frac{dz}{dx}=3x(\frac{1}{x})\)+3logx+5x6 ex+30x5 ex
\(\frac{dz}{dx}\)=3+3logx+5x6 ex+30x5 ex

9. A sphere with the dimensions is shown in the figure. What is the error that can be incorporated in the radius such that the volume will not change more than 2%?

a) 0.06366%
b) 0.006366%
c) 0.6366%
d) 6.366%
View Answer

Answer: b
Explanation: We know that volume of the sphere is,
\(V = \frac{4}{3} πR^3 \)
Differentiating the above equation with respect to R we get,
\(\frac{dV}{dR}= \frac{4}{3} π×3R^2=4πR^2\)
Since the volume of the sphere should not exceed more than 2%,
\(dR=\frac{dV}{4πR^2}=\frac{0.02}{4π(5)^2}=0.00006366\)
Error in radius = 0.006366%

10. Which of the following is correct?
a) \(\frac{d}{dx} (sin^{-1}(⁡x))= \frac{1}{\sqrt{1-x}}\)
b) \(\frac{d}{dx} (sec^{-1}(⁡x))= \frac{1}{x\sqrt{x^2-1}}\)
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:

  • \(\frac{d}{dx} (sin^{-1}⁡(⁡x))= \frac{1}{\sqrt{1-x^2}}\)
  • \(\frac{d}{dx} (sec^{-1}⁡(⁡x))= \frac{1}{x\sqrt{x^2-1}}\)
  • \(\frac{d}{dx} (tan^{-1}(⁡x))= \frac{1}{1+x^2}\)
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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn