## Differentiation Aptitude Quiz

Important topics for Maths has been designed in such a way that it offers very practical and application-based learning to further make it easier for students to understand every concept or topic by correlating it with the day-to-day experiences.

Q1. If x=log⁡p and y=1/p, then
•  (d^2 y)/(dx^2 )-2p=0
•  (d^2 y)/(dx^2 )+y=0
•  (d^2 y)/(dx^2 )+dy/dx=0
•  (d^2 y)/(dx^2 )-dy/dx=0
Solution

Q2.

•  1/(2(1+x)√x)
•  3/((1+x)√x)
•  2/((1+x)√x)
•  3/(2(1+x)√x)
Solution

Q3.  If y=sin⁡x+e^x, then (d^2 y)/(dy^2 )=
•   (-sin⁡x+e^x)^(-1)
•  sin⁡x-e^x /(cos⁡x+e^x)^2
•  sin⁡x-e^x / (cos⁡x+e^x)^3
•  sin⁡x+e^x/(cos⁡x+e^x)^3
Solution

Q4. If sin^(-1)⁡((x^2-y^2)/(x^2+y^2 ))=log⁡a, then dy/dx is equal to
•  x/y
•  y/x^2
•  (x^2-y^2)/(x^2+y^2 )
•  y/x
Solution
(d) We have sin^(-1)⁡((x^2-y^2)/(x^2+y^2 ))=log⁡a ⇒(x^2-y^2)/(x^2+y^2 )=sin⁡(log⁡a) ⇒(1-tan^2⁡Î¸)/(1+tan^2⁡Î¸ )=sin⁡(log⁡a), on putting y=x tan⁡Î¸ ⇒cos⁡2Î¸=sin⁡(log⁡a) ⇒2Î¸=cos^(-1)⁡(sin⁡(log⁡a)) Î¸=1/2 cos^(-1)⁡(sin⁡(log⁡a)) ⇒tan^(-1)⁡(y/x)=1/2 cos^(-1)⁡(sin⁡(log⁡a)) ⇒y/x=tan⁡(1/2 cos^(-1)⁡(sin⁡(log⁡a))) Differentiating w.r.t. x ⇒ (x dy/dx-y)/x^2 =0 ⇒x dy/dx-y=0 ⇒dy/dx=y/x

Q5.

•  (x+(a+b))/√((a-x)(x-b))
•  (2x-a-b)/(2√(a-x) √(x-b))
•  -((a+b))/(2√((a-x)(x-b)))
•  (2x+(a+b))/(2√((a-x)(x-b)))
Solution

Q6. If f(x)=|sin⁡x- |cos⁡x | |, then the value f'(x) at x=7Ï€/6 is
•  Positive
•  (1-√3)/2
• 0
•  None of these
Solution
(a) In the neighbourhood of x=7Ï€/6, we have f(x)=|sin⁡x+cos⁡x |=-sin⁡x-cos⁡x ⇒f^' (x)=-cos⁡x+sin⁡x⇒f'(7Ï€/6)= -cos⁡(7Ï€/6) +sin⁡(7Ï€/6)=(√3-1)/2

Q7.If f(0)=0,f^' (0)=2, then the derivative of y=f(f(f(f(x))) at x=0 is
•  2
•  8
•  16
•  4
Solution
(c) y^' (x)=f^' (f(f(f(x)))) f^' (f(f(x))) f^' (f(x))f'(x) ⇒y^' (0)=f'(f(f(f(0)))) f'(f(f(0)))f'(f(0))f'(0) =f^' (f(f(0))) f^' (f(0)) f^' (0)f'(0) =f^' (f(0)) f^' (0) f^' (0)f'(0) =f^' (0) f^' (0) f^' (0)f'(0) =(f^' (0))^4=2^4=16

Q8.If y=(sin⁡x )^tan⁡x , then dy/dx=
•  (sin⁡x )^tan⁡x (1+sec^2⁡x log sin⁡x )
•  tan⁡x (sin⁡x )^tan⁡x-1 .cos⁡x
•  (sin⁡x )^tan⁡x sec^2⁡x log⁡sin⁡x
•   tan⁡x(sin⁡x )^tan⁡x-1
Solution
(a) y=(sin⁡x )^tan⁡x ⇒log⁡y=tan⁡x log⁡sin⁡x Differentiating w.r.t. x, we get 1/y dy/dx=sec^2⁡x log⁡sin⁡x+tan⁡x 1/sin⁡x .cos⁡x ⇒dy/dx=(sin⁡x )^tan⁡x [1+sec^2⁡x log⁡sin⁡x ]

Q9. If f(x)=sin^(-1)⁡cos⁡x, then the value of f(10)+f'(10) is
•  11-7Ï€/2
•  7Ï€/2-11
•  5Ï€/2-11
•  None of these
Solution

Q10. The nth derivative of xe^x vanishes when
•  x=0
•  x=-1
•  x=-n
• x=n
Solution
(c) f(x)=xe^x f'(x)=e^x+xe^x f''(x)=e^x+e^x+xe^x f'''(x)=2e^x+e^x+xe^x=3e^x+xe^x … … f^n (x)=ne^x+xe^x Now, f^n (x)=0 ⇒ne^x+xe^x=0⇒x=-n

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