How do you use the chain rule to differentiate #y=e^(sinx)#?

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Answer 1 · 2017-01-20T12:02:08

#d/(dx) (e^sinx) = e^sinx cosx#

The chain rule states that:

#d/(dx) f(g(x)) = f'(g(x)) g'(x)#

or, if we pose #u= g(x)#

#(df)/(dx) = (df)/(du)(du)/(dx)#

So in our case if #u=sinx#

#(dy)/(du) = d/(du)e^u = e^u = e^sinx#

#(du)/(dx) = d/(dx) sinx = cosx#

and in conclusion:

#d/(dx) e^sinx = e^sinx cosx#