Question #6b55e

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Answer 1 · 2017-08-23T22:50:08

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

$y = ln(tan^2x)$

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 · 2017-08-24T14:54:54

$dy/dx=2cscxsecx$

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

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 $sin^2x+cos^2x=1$ :

$dy/dx=2cscxsecx$