Question

How do you differentiate `f(x)=sqrttan(2-x^3)` using the chain rule?

Answer 1

`f'(x)=-(3x^2sec^2(2-x^3))/(2sqrt(tan(2-x^3))`

Explanation:

Let `v=-x^3` and `u=tan(2+v)` and `y=f(x)`

Then `y=sqrtu`.

So `f'(x)=dy/dx=dy/(du)*(du)/(dv)*(dv)/(dx)`

`(dv)/dx=-3x^2`

`(du)/(dv)=sec^2(2+v)=sec^2(2-x^3)`

`dy/(du)=1/(2sqrtu)=1/(2sqrt(tan(2-x^3))`

`therefore f'(x)=dy/dx=-3x^2*sec^2(2-x^3)*1/(2sqrt(tan(2-x^3)))`
`=-(3x^2sec^2(2-x^3))/(2sqrt(tan(2-x^3))`

Answer 2

`f'(x)=(-3x^2sec^2(2-x^3))/(2sqrt(tan(2-x^3))`

Explanation:

We will use the chain rule where we take the derivative of the outside and multiply it by the derivative of the inside.

We have `f(x)=sqrt(tan(2-x^3)` It's better to rewrite it as an exponent `(tan(2-x^3))^(1/2)`

So, first lets take the derivative of the `color(blue)("outside")` using the power rule:

`color(blue)(1/2)tan(2-x^3)^color(blue)(1/2-2/2)=> 1/2tan(2-x^3)^(-1/2)`

This is just one part of the answer.

Now let's take the derivative of the `color(green)("inside")`

`1/2(color(green)tan(2-x^3))^(-1/2)`

`d/dxcolor(green)tan(2-x^3) => sec^2(2-x^3)(-3x^2)`

We ended up using the chain rule on the `color(green)("inside")` too.

The derivative of the `tanx` is `sec^2x` but we also have to apply the chain rule here too because `sec^2x` has a function in the inside too. Take the derivative of the `color(red)("function")` inside `sec^2color(red)((2-x^3))` which is where we got the `-3x^2` from.

Now we just multiply both the derivatives of the outside and inside.

`1/(2sqrt(tan(2-x^3)))xxsec^2(2-x^3)(-3x^2)`

`d/dxsqrt(tan(2-x^3)` `=` `(-3x^2sec^2(2-x^3))/(2sqrt(tan(2-x^3))`