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How do you find the derivative of #y=sin(tan2x)#?

Answer 1

I would use the Chain rule deriving #sin# first and then #tan# as: #y'=cos(tan(2x))*1/cos^2(2x)*2=(2cos(tan(2x)))/cos^2(2x)#

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Answer 2

Axel Alvarez

To find the derivative of ( y = \sin(\tan^2(x)) ), we can use the chain rule.

Let ( u = \tan^2(x) ). Then ( y = \sin(u) ).

The derivative of ( u ) with respect to ( x ) is ( \frac{d}{dx}(\tan^2(x)) = 2\tan(x)\sec^2(x) ).

Now, using the chain rule, the derivative of ( y ) with respect to ( x ) is:

[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} ]

[ = \cos(u) \cdot 2\tan(x)\sec^2(x) ]

[ = 2\sin(\tan^2(x))\tan(x)\sec^2(x) ]

So, the derivative of ( y = \sin(\tan^2(x)) ) with respect to ( x ) is ( 2\sin(\tan^2(x))\tan(x)\sec^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.