Question

`y=sin^2(lnx^2)` What is `dy/dx` ?

`y=sin^2(lnx^2)` . Find `dy/dx`

Answer 1

`dy/dx = (4sin(lnx^2)cos(lnx^2))/x`

Explanation:

If `y=sin^2(lnx^2)` then

`dy/dx=d/(dlnx^2)sin^2(lnx^2) d/dxlnx^2=2sin(lnx^2)cos(lnx^2)(2x)/x^2=(4sin(lnx^2)cos(lnx^2))/x`

Answer 2

`dy/dx=(4sin(ln(x^2))cos(ln(x^2)))/x`

Explanation:

We use the chain rule twice here:

`d/dx[f(g(x))]=f'(g(x))*g'(x)`

`=>dy/dx=d/dx[sin^2(ln(x^2))]`

Some rules to recall here:

`d/dx[x^n]=nx^(n-1)` if `n` is a constant.

`d/dx[ln(x)]=1/x`

`d/dx[sinx]=cosx`

`=>dy/dx=2*sin(ln(x^2))*d/dx[sin(ln(x^2))]`

`=>dy/dx=2*sin(ln(x^2))*cos(ln(x^2))*d/dx[ln(x^2)]`

`=>dy/dx=2*sin(ln(x^2))*cos(ln(x^2))*1/(x^2)*d/dx[x^2]`

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

`=>dy/dx=4*sin(ln(x^2))*cos(ln(x^2))*1/(x)`

`=>dy/dx=(4sin(ln(x^2))cos(ln(x^2)))/x`

That is our answer!