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

1 Answer
Steve M
Nov 10, 2016

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

Explanation:

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 #

Related questions
See all questions in Chain Rule
Impact of this question
41723 views around the world
You can reuse this answer
Creative Commons License