Question

y=2sin²x find dy/dx​

Answer 1

Answer:

2sin2x

Step-by-step explanation:

using multiplication formula of differentiation

y = udv/dt + vdu/dt

y = 2d(sin^2x)/dt + sin^2d(2)/dt

y = 2sin2x + sin^2 * 0

y = 2sin2x

thank u

hope it helps :)

Answer 2

Solution :-

Given, y = 2sin²x

We have to find dy/dx.

Formulas to be used :

→ d/dx (xⁿ) = nx^(n - 1)

→ d/dx (sinx) = cosx

→ 2sinxcosx = sin2x

Now,

→ dy/dx = 2 d/dx (sin²x)

→ dy/dx = 2 d/dx (sinx)²

→ dy/dx = 2 × 2sinx d/dx(sinx)

→ dy/dx = 2 × 2sinx cosx

→ dy/dx = 2sin2x

Therefore,

Required value of dy/dx = 2sin2x.