Mathematics Questions and Answers – Differentiability

«
»

This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Differentiability”.

1. Find the derivative of f(x) = sin(x2).
a) -sin(x2)
b) 2xcos(x2)
c) -2xcos(x2)
d) -2xsin(x2)
View Answer

Answer: b
Explanation: Differentiation of the function f(x) = sin(x2) is done with chain rule. First we differentiate sin function which becomes cos and then differentiate the inner (x2) which becomes 2x, hence it comes out to be 2xcos(x2).
advertisement

2. What is derivative of xn?
a) n
b) nxn
c) nxn-1
d) nxn-2
View Answer

Answer: c
Explanation: It is a standard rule for derivative of a function of this form in which the original power comes in front and the value in the power is decreased by one. Therefore the only option of this form is nxn-1 from the given options.

3. Find derivative of tan(x+4).
a) sec2(x+4)
b) 4 sec2(x+4)
c) 4x sec2(x+4)
d) sec2(x)
View Answer

Answer: a
Explanation: We know that derivative of tanx is sec2(x), now in the above question we get tan(x+4), hence its derivative comes out to be sec2(x+4), as the inside expression (x+4) is differentiated into 1.
Therefore the answer is sec2(x+4).
advertisement
advertisement

4. What is value of \(\frac{dy}{dx}\) if x-y = 1?
a) 1
b) 2
c) -1
d) 2
View Answer

Answer: a
Explanation: We know x-y = 1, hence we differentiate it on both sides-:
We get 1- \(\frac{dy}{dx}\) = 0, \(\frac{dy}{dx}\) = 1, hence the value of \(\frac{dy}{dx}\) comes out to be 1.

5. Value after differentiating cos (sinx) is _________
a) sin (sinx).cosx
b) -sin (sinx).cosx
c) sin (sinx)
d) sin (cosx).cosx
View Answer

Answer: b
Explanation: We differentiate the given function with the help of chain rule so we first differentiate the outer function which becomes –sin and then we differentiate the inner function sinx which is differentiated and comes out to be cosx, hence the differentiated function comes out to be -sin (sinx).cosx.
advertisement

6. Value after differentiating cos (x2+5) is ________
a) 5.sin (x2+5)
b) -sin (x2+5).2x
c) sin (x2+5).2x
d) cos (x2+5).2x
View Answer

Answer: b
Explanation: We differentiate the given function with help of chain rule and hence the outer function becomes –sin and the inner function is differentiated into 2x, therefore the answer comes out to be -sin (x2+5).2x.

7. Find \(\frac{dy}{dx}\) of 2x+3y = sinx.
a) \(\frac{cosx-2}{3}\)
b) \(\frac{cosx-2}{2}\)
c) \(\frac{cosx-3}{2}\)
d) \(\frac{sinx-2}{3}\)
View Answer

Answer: a
Explanation: Differentiating on both sides we get 2 + 3\(\frac{dy}{dx}\) = cosx.
3\(\frac{dy}{dx}\) = cosx-2.
\(\frac{dy}{dx} = \frac{cosx-2}{3}\).
advertisement

8. What is derivative of cotx?
a) tanx
b) –sec2x
c) –cosec2x
d) cosec2x
View Answer

Answer: c
Explanation: The derivative of cotx is –cosec2x, as this function has a fixed derivative like sinx has its derivative cosx. Therefore the answer to the above question is –cosec2x.

9. If \(y = tan^{-1}(\frac{3x-x^3}{1-3x^2}), \frac{-1}{\sqrt{3}} < x < \frac{-1}{\sqrt{3}}\)
a) 3
b) \(\frac{3}{1+x}\)
c) –\(\frac{3}{1+x^2}\)
d) \(\frac{3}{1+x^2}\)
View Answer

Answer: d
Explanation: Given function is \(y = tan^{-1}(\frac{3x-x^3}{1-3x^2})\),
Now taking RHS and substituting x = tang in it and then we get,
y = \(tan^{-1}(\frac{3tang-tang^3}{1-3tang^2})\), Now it becomes the expansion of the function tan3g,
Hence the given function becomes y= tan-1(tan3g). Which is equal to 3g, now substituting the value of g= tan-1x, now after differentiating both sides we get the answer \(\frac{3}{1+x^2}\).
advertisement

10. Find \(\frac{dy}{dx}\) of y = sin (ax + b).
a) a.cos (ax + b)
b) b.sin (ax + b)
c) a.sin (ax + b)
d) a.cos (ax + b)
View Answer

Answer: a
Explanation: We differentiate the given function with the help of chain rule and hence
the outer function is differentiated into cos, and the inner function comes out to be a and the constant b becomes 0, which is multiplied to the whole function and the answer comes out to be
=> a.cos (ax + b).

Sanfoundry Global Education & Learning Series – Mathematics – Class 12.

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

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!

advertisement
advertisement
Manish Bhojasia - Founder & CTO at Sanfoundry
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 | Youtube | Instagram | Facebook | Twitter