Question

derivative of sin square (ax+b)​

Answer 1

AnswEr :

Given that,

$\sf \: y = sin {}^{2} (ax + b)$

We have to find the derivative of y .

The given expression can be rewritten as,

$\sf \: y = sin(ax \: + b).sin(ax + b)$

The above expression is of the form uv

Derivative of y would be of the form :

$\sf \: y' = u'v + v'u$

This is known as Product Rule of Differentiation.

Now,

$\longrightarrow \sf y' = \dfrac{d \bigg(sin(ax + b) \bigg)}{dx} sin(ax + b) + \dfrac{d \bigg(sin(ax + b) \bigg)}{dx} sin(ax + b)$

We would have to apply Chain Rule eventually.

• Derivative of sin(ax + b) is cos(ax + b)

$\longrightarrow \: \sf \: y' = sin(ax + b)cos(ax + b) \times \dfrac{d(ax + b)}{dx} + sin(ax + b)cos(ax + b). \dfrac{d(ax + b)}{dx} \\ \\ \longrightarrow \: \sf \: y' = 2a \: sin(ax + b)cos(a + b)$

Here,

2sin(ax +b)cos(ax + b) = sin2(ax + b)

Thus,

$\longrightarrow \: \boxed{ \boxed{ \sf y' = a.sin2(ax + b)}}$

Derivative of the above expression is a.sin2(ax + b)