What is the derivative of #y=sin(tan2x)#?

To find the derivative of ( y = \sin(\tan^2(x)) ), we can use the chain rule. The chain rule states that if we have a composition of functions, ( f(g(x)) ), then the derivative is ( f'(g(x)) \cdot g'(x) ).

So, for our function ( y = \sin(\tan^2(x)) ), let's denote ( u = \tan^2(x) ). Then, ( y = \sin(u) ).

Now, taking the derivative of ( y = \sin(u) ) with respect to ( u ), we get ( \frac{dy}{du} = \cos(u) ).

Next, taking the derivative of ( u = \tan^2(x) ) with respect to ( x ), we get ( \frac{du}{dx} = 2\tan(x)\sec^2(x) ).

Finally, applying the chain rule, we have:

[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \cos(u) \cdot 2\tan(x)\sec^2(x) ]

Substituting ( u = \tan^2(x) ), we get:

[ \frac{dy}{dx} = \cos(\tan^2(x)) \cdot 2\tan(x)\sec^2(x) ]

So, the derivative of ( y = \sin(\tan^2(x)) ) is ( \cos(\tan^2(x)) \cdot 2\tan(x)\sec^2(x) ).