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

Answer 1

Ezekiel Alexander

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

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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:

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#

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Answer 2

Samantha Adkison

To find the derivative of ln((tan^2)x), you can use the chain rule and the derivative of ln(u) which is (1/u) * u'.

Let y = ln((tan^2)x).
Apply the chain rule: dy/dx = (1/(tan^2)x) * (2tan(x) * sec^2(x)).
Simplify: dy/dx = 2sec^2(x).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.