Question

Question #e662a

Answer 1

`dy/dx=(2xcos(x^2))/sin(x^2)`

Explanation:

To calculate the derivative of `log(sin(x^2))` we will use the chain rule.

The chain rule is #(color(red)V(color(orange)U(color(purple)x)))^'=color(red)V^'(color(orange)U(color(purple)x))*color(orange)U^'(color(purple)x) * (color(purple)x)^'#

`y=color(red)logcolor(orange)(sin(color(purple)x^color(purple)2))`

`dy/dx=d/dxcolor(red) log(color(orange) sin( color(purple)x^color(purple)2))*d/dxcolor(orange) sin(color(purple)x)^color(purple)2*d/dx(color(purple)(x)^color(purple)2)`

`dy/dx=1/sin(x^2)*cos(x^2)*2x`

`dy/dx=(2xcos(x^2))/sin(x^2)`

Answer 2

Let's see.

Explanation:

Given, `y=log(sinx^2)`

Now, differentiating w.r.t `x` and applying chain rule:

`dy/dx=d/dx(log(sinx^2))`

`:.dy/dx=1/(sinx^2)xxd/dx(sinx^2)`

`:.dy/dx=1/(sinx^2)xx(cosx^2)xxd/dx(x^2)`

`:.dy/dx=(cosx^2/sinx^2)xx(2x)`

`:.dy/dx=2xcdotcotx^2`. (Answer).

Hope it Helps:)