Question

Let u be a differentiable function of x, in exercise 123, the following result is proved.

Answer 1

We need to find the derivative of the following equation:

`h(x)=|9x|\cos (2x)`

Since it's the product of two functions, we need to apply the following expression:

`\begin{gathered} h(x)=f(x)\cdot g(x) \\ h^(\prime)(x)=f^(\prime)(x)\cdot g(x)+f(x)\cdot g^(\prime)(x) \end{gathered}`

Therefore we have:

`\begin{gathered} h^(\prime)(x)=9\cdot(9x)/(|9x|)\cdot\cos (2x)-|9x|\cdot2\cdot\sin (2x) \\ h^(\prime)(x)=9\cdot(9x)/(9\cdot|x|)\cdot cos(2x)-|9x|\cdot2\cdot\sin (2x) \\ h^(\prime)(x)=(9x)/(|x|)\cdot\cos (2x)-|9x|\cdot2\cdot\sin (2x) \\ h^(\prime)(x)=\frac{9x\lbrack\cos (2x)-2x\cdot\sin (2x)\rbrack_{}}x \end{gathered}`