Question

How do you find the derivative of `(sinx^2)/(tanx)`?

Answer 1

`(2xcos(x^2)tanx-sec^2xsin(x^2))/tan^2x`

Explanation:

According to the quotient rule, `(f/g)'=(f'g-fg')/g^2`

Here, `f=sin(x^2)` and `g=tan(x)`

`d/dx(sinx^2/tanx)=(d/dx(sin(x^2))*tanx-d/dx(tanx)*sin(x^2))/(tanx)^2`

We first compute:

`d/dx(sin(x^2))`

If we take the chain rule, where `f(u)=sin(x^2)`, we have `f=sin(u)` and `u=x^2`

So we have:

`d/(du)sin(u)*d/dxx^2`

`cosu*2x`

And as `u=x^2`, we have:

`d/dx(sin(x^2))=2xcos(x^2)`

Then ,we compute:

`d/dxtanx=sec^2(x)`

So we can input into our original equation:

`(2xcos(x^2)tanx-sec^2xsin(x^2))/tan^2x`

The answer.

Answer 2

`2xcosx^2cotx-sinx^2csc^2x`

Explanation:

Let `f(x)=sinx^2`, and `g(x)=tanx`

The formula is

`(f/g)^'=(f^'g-g^'f)/g^2`

The derivative of `sinx` is `cosx`, but here we have `sinx^2` so the derivative will be `coscolor(red)(x^2` multiplied by the derivative of `color(red)(x^2` which is `2x`, so `2xcosx^2`

So `color(blue)(f^'=2xcosx^2`

`g^'` is the derivative of `tanx` which is `sec^2x`

So `color(green)(g^'=sec^2x`

Substituting these values in we get:-

`(f/g)^'=(color(blue)(2xcosx^2)tanx-color(green)(sec^2x)sinx^2)/color(brown)(tan^2x)`

Now we just simplify, separate the denominator:-

`(2xcosx^2)/tanx-(color(green)((1/cos^2x))sinx^2)/color(brown)(sin^2x/cos^2x)`

`=2xcosx^2cotx-sinx^2/cancel(cos^2x)*cancel(cos^2x)/sin^2x`

`=2xcosx^2cotx-sinx^2csc^2x`