Question

Question #6b55e

Answer 1

I will assume that we want to find `dy/dx`

Explanation:

`y = ln(tan^2x)`

Use `d/dx(lnu) = 1/u (du)/dx` (chain rule)

And the chain rule again

`d/dx(tan^2u) = d/dx((tanx)^2) = 2tanx d/dx(tanx) = 2tanxsec^2x`

So,

`dy/dx = 1/(tan^2x) d/dx(tan^2x)`

`= cot^2x (2tanxsec^2x)`

`= 2cotxsec^2x`

`= 2cscxsecx`

Answer 2

`dy/dx=2cscxsecx`

Explanation:

We can also simplify the function using the following logarithm rules:

• `log(a^b)=blog(a)`
• `log(a/b)=log(a)-log(b)`

So:

`y=ln(tan^2x)=2ln(tanx)=2ln(sinx/cosx)=2ln(sinx)-2ln(cosx)`

Taking the derivative, we need to remember that `d/dxlnx=1/x`. We also need to use the chain rule:

`dy/dx=2/sinx(d/dxsinx)-2/cosx(d/dxcosx)`

`dy/dx=(2cosx)/sinx+(2sinx)/cosx=(2(cos^2x+sin^2x))/(sinxcosx)`

Since `sin^2x+cos^2x=1`:

`dy/dx=2cscxsecx`