Question

How do you differentiate `y = tan^3(x^3)`?

Answer 1

Use the chain rule twice.

Explanation:

It is important to remember the convention for powers of trigonometric functions:

`tan^3(x^3)` means "raise `x` to the third power, then find the tangent of that number and finally raise that tangent to the third power". In notation:

`(tan(x^3))^3`

The outermost function is the (second) cube.
And we know that `d/dx (u^3) = 3u^2 (du)/dx`
In this case `u = tan(x^3)`

So we get

`dy/dx = 3(tan(x^3))^2 * d/dx(tan(x^3))`

Now we need the derivative of `tan(x^3)`.

We'll use the chain rule again:

`d/dx(tanu) = sec^2u (du)/dx`

We get

`d/dx(tan(x^3)) = sec^2(x^3)* d/dx(x^3)`

`= sec^2(x^3)* 3x^2`

Putting it all together, we have:

`y = tan^3(x^3)`

`dy/dx = 3tan^2(x^3)*sec^2(x^3)*3x^2`

Which can be written:

`dy/dx = 9x^2tan^2(x^3)sec^2(x^3)`