How do you differentiate # tan(sin x)#?

Answer 1

Paisley Johnson

#sec^2(sinx)cosx#

Given: #d/dx(tan(sinx))#.

Here, let #y=tan(sinx)#.

Use the chain rule, which states that,

#dy/dx=dy/(du)*(du)/dx#

Let #u=sinx,:.(du)/dx=cosx#.

Then #y=tanu,dy/(du)=sec^2u#.

So, combining our results, we get:

#dy/dx=sec^2u*cosx#

Substituting back #u=sinx#, we get:

#dy/dx=sec^2(sinx)*cosx#

#=sec^2(sinx)cosx#

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Answer 2

William Abdalla

To differentiate tan(sin x), you use the chain rule, which states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function. So, the derivative of tan(sin x) is sec^2(sin x) * cos 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.