## Differentiation 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.

•  Equal to 0
•  Equal to 1
•  In π/2
•  Non-existent
Solution
(a) Let g(x)=(sin⁡x )^(In x)=e^(In x.In(sin⁡x) ) f(x)=g^' (x)=(sin⁡x )^(In x) [cot⁡x(In x)+(In(sin⁡x))/x] Hence, f(π/2)=g^' (π/2)=1(0+0)=0

Q2.If y^2=P(x), a polynomial of degree 3, then 2 d/dx (y^2 (d^2 y)/(dx^2 ))=
•  P'''(x)+P'(x)
•  P''(x)+P'''(x)
•  P(x)+P'''(x)
•  A constant
Solution
(c) We have y^2=P(x), where P(x) is a polynomial of degree 3 and hence thrice differentiable, Then y^2=P(x) (1) Differentiate(1) w.r.t. x, we get 2y dy/dx=P'(x) (2) Again differentiate w.r.t. x, we get

Q3.  If y=log_sin⁡x (tan⁡x ), then (dy/dx)_(π/4) is equal to
•   4/log⁡2
•  -4 log 2
•  (-4)/log⁡2
•  None of these
Solution
(c) y=log⁡tan⁡x /log⁡sin⁡x ⇒ dy/dx=((log⁡sin⁡x) ((sec^2 x)/tan⁡x )-(log⁡tan⁡x )(cot⁡x ) )/(log⁡sin⁡x )^2 ⇒(dy/dx)_(π/4)=(-4)/log⁡2 (On simplification)

Q4. If y=ae^mx+be^(-mx), then (d^2 y)/(dx^2 )-m^2 y is equal to
•  m^2 (ae^mx-be^(-mx))
•  1
•  0
•  None of these
Q5. If y=tan^(-1)⁡(2^x/(1+2^(2x+1) )), then dy/dx at x=0 is
•  1
•  2
•  In 2
•  None of these
Q6. If (sin⁡x)(cos⁡y )=1/2, then d^2 y/dx^2 at (π/4,π/4) is
•  -4
•  -2
• -6
•  0
Solution
(a) (sin⁡x )(cos⁡y )=1/2 (cos⁡x )(cos⁡y )-sin⁡y sin⁡x dy/dx=0 ⇒dy/dx=(cot⁡x)(cot⁡y) ⇒(d^2 y)/(dx^2 )= -cosec^2⁡x.cot⁡y - cosec^2⁡y cot⁡x.dy/dx Now (dy/dx)_((π/4,π/4))=1 ⇒((d^2 y)/(dx^2 ))_((π/4,π/4))=-(2)(1)-(2)(1)(1)=-4

Q7.If u=x^2+y^2 and x=s+3t,y=2s-t, then (d^2 u)/(ds^2 ) equals to
•  12
•  32
•  36
•  10
Q8.If y=sec⁡(tan^(-1)⁡x ), then dy/dx at x=1 is
•  cos⁡π/4
•  sinπ/2
•  sin⁡π/6
•  cos⁡π/3
Q9.If y=sin^(-1)⁡x/√(1-x^2), then (1-x^2 ) dy/dx is equal to
•  x+y
•  1+xy
•  1-xy
•  xy-2
Solution
(b) y=sin^(-1)⁡x/√(1-x^2) dy/dx=(√(1-x^2) 1/√(1-x^2)-(sin^(-1)⁡x) 1/2 ((-2x))/√(1-x^2))/(1-x^2) ⇒(1-x^2) dy/dx=1+x(sin^(-1)⁡x/√(1-x^2 ))=1+xy

Q10. Let h(x) be differentiable for all x and let f(x)=(kx+e^x )h(x), where k is some constant. If h(0)=5,h^'(0)=-2 and f^'(0)=18, then the value of k is
•  5
•  4
•  3
• 2.2)
Solution
(c) f^' (x)=(kx+e^x ) h^' (x)+h(x)(k+e^x) f^' (0)=h^' (0)+h(0)(k+1) ⇒18= -2+5(k+1)⇒k=3 #### Written by: AUTHORNAME

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