Question

What is the derivative of `sinx^tanx`?

Answer 1

`cos(x^(tan(x)))x^(tan(x))(tan(x)/x+ln(x)*sec^2(x))`
[Assuming you meant `sin(x^(tan(x)))`]

Explanation:

We have:

`d/dx[sin(x^(tan(x)))]` We use the chain rule:

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

Also remember that `d/dx[sin(x)]=cos(x)`

`=>cos(x^(tan(x)))*d/dx[x^(tan(x))]`

Here is a rule that you may not be familiar with:

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

`=>cos(x^(tan(x)))*x^(tan(x))*d/dx[ln(x)*tan(x)]`

Use the product rule:

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

`=>cos(x^(tan(x)))x^(tan(x))(d/dx[ln(x)]*tan(x)+ln(x)*d/dx(tan(x)))`

Remember that:

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

`d/dx(tan(x))=sec^2(x)`

`=>cos(x^(tan(x)))x^(tan(x))(1/x*tan(x)+ln(x)*sec^2(x))`

`=>cos(x^(tan(x)))x^(tan(x))(tan(x)/x+ln(x)*sec^2(x))`

Answer 2

`(sinx)^(tanx)*(((cosx)(tanx))/(sinx)+ln(sinx)*sec^2x)`
[Assuming you meant `(sinx)^(tanx)`]

Explanation:

We have:

`d/dx[(sinx)^(tanx)]`

We use a rule you may be unfamiliar with:

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

Therefore, we have:

`(sinx)^(tanx)*d/dx[ln(sinx)*tanx]`

The product rule:

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

`=>(sinx)^(tanx)*(d/dx[ln(sinx)]*tanx+ln(sinx)*d/dx[tanx])`

Remember that:

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

`d/dx(lnx)=1/x`

`d/dx(tanx)=sec^2x`

`=>(sinx)^(tanx)*(1/(sinx)*d/dx(sinx)*tanx+ln(sinx)*sec^2x)`

Another thing to remember here:

`d/dx(sinx)=cosx`

`=>(sinx)^(tanx)*(1/(sinx)*cosx*tanx+ln(sinx)*sec^2x)` Simplify.

`=>(sinx)^(tanx)*(((cosx)(tanx))/(sinx)+ln(sinx)*sec^2x)`

This is our answer!