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

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Answer 1 · 2018-04-28T14:53:15

#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#