How do you differentiate #f(x)=e^cosx#?

Question

Emma Aldridge · Accepted Answer

# f'(x) = -sinxe^cosx #

f you are studying maths, then you should learn the Chain Rule for Differentiation, and practice how to use it:

If # y=f(x) # then # f'(x)=dy/dx=dy/(du)(du)/dx #

I was taught to remember that the differential can be treated like a fraction and that the "#dx#'s" of a common variable will "cancel" (It is important to realise that #dy/dx# isn't a fraction but an operator that acts on a function, there is no such thing as "#dx#" or "#dy#" on its own!). The chain rule can also be expanded to further variables that "cancel", E.g.

# dy/dx = dy/(dv)(dv)/(du)(du)/dx # etc, or # (dy/dx = dy/color(red)cancel(dv)color(red)cancel(dv)/color(blue)cancel(du)color(blue)cancel(du)/dx) #

So Let # y = f(x) = e^(cosx) #, Then:

# { ("Let "u=cosx, => , (du)/dx=-sinx), ("Then "y=e^u, =>, dy/(du)=e^u ) :}#

Using # dy/dx=(dy/(du))((du)/dx) # we get:

# dy/dx = (e^u)(-sinx) # # :. dy/dx = -sinxe^cosx #

Emily Ammerman · Answer

undefined To differentiate the function $ f(x) = e^{\cos(x)} $, you would use the chain rule. The chain rule states that if you have a composition of functions, the derivative is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function.

So, for $ f(x) = e^{\cos(x)} $:

$$ f'(x) = \frac{d}{dx} \left( e^{\cos(x)} \right) = e^{\cos(x)} \times (-\sin(x)) $$

Therefore, the derivative of $ f(x) = e^{\cos(x)} $ is $ -e^{\cos(x)}\sin(x) $.