Test: Differentiability And Chain Rule - JEE MCQ

y = V^3/3
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

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

y = log(secx + tanx)
dy/dx = 1/(secx + tanx){(secxtanx) + sec^2x}
= secx(secx + tanx)/(secx + tanx)
= secx

Given: y = sin 3x . sin³

= 3 sin²x.sin 4x

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

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

 Concept:

Calculation:

Hence, option (c) is correct.