how to take a derivative of a sin^2(x) function

Question

So I looked over the book and now I know the derivatives of the basic trig functions but I don't know what sin^2 would be. If you just square the regular derivative would

sin^2(8x) − πx have the derivative cos^2(8x)*8- Π ? I don't know how to do this other than that method. Could you please explain what to do.

 

Answer 1

Let u = sin x

 

Now, you have u² = sin²(x)

 

The derivative of u² = 2u du

 

If u = sin x, then du = cos x dx

 

So the derivative of sin²(x) = 2 sin x cox dx 

 

The derivative of sin²(8x) = 8[2 sin(8x) cos(8x) dx] by the Chain Rule 

 

= 16 sin (8x) cos (8x) dx 

 

Of course, the derivative of (pi)x = pi

 

The entire derivative of the given function is 16 sin (8x) cos (8x) - pi

Answer 2

Use the chain rule:

 

d/dx [sin²(8x)] = 2 sin(8x) d/dx [sin(8x)] = 2 sin(8x) cos(8x) * d/dx (8x)

 

= 2 sin(8x) cos(8x) * 8

 

= 16 sin(8x) cos(8x)

 

I think you can finish the solution from here.