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

Question

Cameron Alexander · Accepted Answer

#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#

Luke Alvarez · Answer

undefined To differentiate $ y = e^{\sin(x)} $ using the chain rule:

1. Recognize that $ y $ is a composition of two functions: the outer function is $ e^u $ and the inner function is $ \sin(x) $.
2. Apply 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.
3. Derivative of the outer function $ e^u $ with respect to $ u $ is $ e^u $.
4. Derivative of the inner function $ \sin(x) $ with respect to $ x $ is $ \cos(x) $.
5. Combine these derivatives to get the derivative of $ y = e^{\sin(x)} $ with respect to $ x $:
   $$ \frac{dy}{dx} = e^{\sin(x)} \cdot \cos(x) $$
   or, equivalently,
   $$ \frac{dy}{dx} = \cos(x) \cdot e^{\sin(x)} $$