Bader,

I think you want to find dy/dx.

This is an exercise in using the chain rule. Let's try a slightly different problem.

Find dy/dx if y = tan(x² + 1)

This function is composed of two functions, the function f(s) = tan(s) and g(s) = s² + 1. I have used the variable s here rather than x because the variable is only used to help describe how the function operates. f is the tangent function and g is the function "square the variable and add 1". The function y is constructed from f and g by

y = f(g(x))

that is

y = f(x² + 1) = tan(x² + 1)

The chain rule says that

dy/dx = f '(g(x)) g'(x)

The derivative of the tangent function is the square of the secant function so

dy/dx = sec²(g(x)) g'(x)

But g(x) = x² + 1 and g'(x) = 2x so

dy/dx = sec²(x² + 1) 2x

Now try f(x) = sin(2x)

Penny