Question

How do you find the derivative of `ln((tan^2)x)`?

Answer 1

`2cotx+2tanx` or `2secxcscx`, depending on preferred simplification

Explanation:

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Method 1 - No simplification

`y=ln(tan^2x)`

Note that `d/dxln(x)=1/x`, so by the chain rule `d/dxln(u)=1/u*(du)/dx`. Then:

`dy/dx=1/tan^2x*d/dxtan^2x`

And we need to use the chain rule again since we have a function squared:

`dy/dx=1/tan^2x * 2tanx * d/dxtanx`

`dy/dx=1/tan^2x * 2tanx * sec^2x`

`dy/dx=cos^2x/sin^2x * (2sinx)/cosx * 1/cos^2x`

`dy/dx=2/(sinxcosx)`

`dy/dx=2secxcscx`

Method 2 - Simplification

Use the logarithm rules:

• `ln(a^b)=bln(a)`
• `ln(a//b)=ln(a)-ln(b)`

Then:

`y=ln(tan^2x)`

`y=ln(sin^2x/cos^2x)`

`y=ln(sin^2x)-ln(cos^2x)`

`y=2ln(sinx)-2ln(cosx)`

Then differentiating becomes easier:

`dy/dx=2* 1/sinx * d/dxsinx-2 * 1/cosx * d/dxcosx`

`dy/dx=2(cosx/sinx+sinx/cosx)`

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

`dy/dx=2/(sinxcosx)`

`dy/dx=2secxcscx`