Test: Differentiability And Chain Rule - JEE MCQ

# Test: Differentiability And Chain Rule - JEE MCQ

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## 10 Questions MCQ Test Mathematics (Maths) for JEE Main & Advanced - Test: Differentiability And Chain Rule

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Test: Differentiability And Chain Rule - Question 1

### The dervative of sin(x2) is

Test: Differentiability And Chain Rule - Question 2

### If   where V = x2 - 2x + 3 then dy/dx is

Detailed Solution for Test: Differentiability And Chain Rule - Question 2

y = V3/3
V = x2 - 2x + 3
y = 1/3*v3
dy/dx = d/dx{1/3*(x2-2x+3)3
= 1/3*3*(x2-2x+3)2*(2x-2)
{d/dx(xn)=nx^n-1*dx/dx}
= 2(x-1)*(x2-2x+3)2

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Test: Differentiability And Chain Rule - Question 3

### F(x) = tan (log x) F'(x) =

Detailed Solution for Test: Differentiability And Chain Rule - Question 3

∫tan(log x) dx
log x = t
x = et
dx = et dt
f(t) = ∫tan t dt
f’(t)= sec2 t
= sec2(log x)

Test: Differentiability And Chain Rule - Question 4

Which of the following function is not differentiable at x = 0?​

Test: Differentiability And Chain Rule - Question 5

Test: Differentiability And Chain Rule - Question 6

y = log(sec + tan x)

Detailed Solution for Test: Differentiability And Chain Rule - Question 6

y = log(secx + tanx)
dy/dx = 1/(secx + tanx){(secxtanx) + sec2x}
= secx(secx + tanx)/(secx + tanx)
= secx

Test: Differentiability And Chain Rule - Question 7

If y=sin 3x . sin3 x then dy/dx is

Detailed Solution for Test: Differentiability And Chain Rule - Question 7

Given: y = sin 3x . sin3

= 3 sin2x.sin 4x

Test: Differentiability And Chain Rule - Question 8

Detailed Solution for Test: Differentiability And Chain Rule - Question 8

LHD = limh→0 ((−h)2 sin(−1/h) − 0)/−h
limh→0 −h2 sin(1/h)/−h
= limh→0 h × sin(1/h)
= 0
RHD = limh→0 (h2 sin(1/h) − 0)/h
= limh→0 h × sin(1/h)
= 0
RHD=LHD
∴ f(x) is differentiable at x= 0

Test: Differentiability And Chain Rule - Question 9

​Find the derivate of y = sin4x + cos4x

Detailed Solution for Test: Differentiability And Chain Rule - Question 9

y=sin4x,  and z=cos4x
So by using chain rule
df(x)/dx = dsin4x/dx + dcos4x/dx
=dy4/dy * dy/dx + dz4/dz * dzdx
=dy4/dy * dsinxdx + dz4/dz * dcosx/dx
=4y(4−1)⋅cosx+4z(4−1)⋅(−sinx)
=4sin3xcosx − 4cos3xsinx
=4sinxcosx(sin2x − cos2x)
=2sin2x(−cos2x)
=−2sin2xcos2x
=−sin4x

Test: Differentiability And Chain Rule - Question 10

If  is equal to

Detailed Solution for Test: Differentiability And Chain Rule - Question 10

Concept:

Calculation:

Hence, option (c) is correct.

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## Mathematics (Maths) for JEE Main & Advanced

209 videos|443 docs|143 tests