How do you differentiate #f(x)=e^(tansqrtx)# using the chain rule.?

1 Answer
Apr 12, 2016

Answer:

#f'(x) = e^tan(sqrtx) * sec^2sqrtx/(2sqrtx)#

Explanation:

Chain rule is basically a lot of substitution.

let #u=tan(sqrtx)#

now #f(x) = e^u#

so if we try to take the derivative now we apply chain rule to get:
#f'(x) = e^u * (du)/dx#

so now we are on a mission to find #(du)/dx#...

We are going to need to rewrite our #u# equation to something like this:
#u=tan(w)#
#w=sqrtx#

Derive #u#

#(du)/dx = sec^2w * (dw)/dx#

Now to find #(dw)/dx#...

#w=sqrtx#
Yay we know how to do this one!

#(dw)/dx= 1/2x^(-1/2)=1/(2sqrtx)#

now substitute back into #(du)/dx#

#(du)/dx = sec^2w * 1/(2sqrtx) = sec^2w/(2sqrtx)#

and finally substitute back into #f'(x)#

#f'(x) = e^u * sec^2w/(2sqrtx)#

now change everything back to terms of x (you could have done this step earlier)

#f'(x) = e^tan(sqrtx) * sec^2sqrtx/(2sqrtx)#