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

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 with # y = f(x) = e^(cot sqrt(x)) #, Then:

# { ("Let",u=sqrt(x)=x^(1/2), => , (du)/dx=1/2x^(-1/2)=1/(2sqrt(x))), ("And",v=cot sqrt(x)=cotu, => , (dv)/(du)=-csc^2u), ("Then",y=e^(cot sqrt(x))=e^v, =>, dy/(dv)=e^v ) :}#

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

# \ \ \ \ \ dy/dx = (e^v)(-csc^2u)(1/(sqrt(x))) #
# \ \ \ \ \ \ \ \ \ \ = (e^(cot sqrt(x)))(-csc^2sqrt(x))(1/(2sqrt(x))) #
# \ \ \ \ \ \ \ \ \ \ = -(csc^2(sqrt(x)) * e^(cot sqrt(x))) / (2sqrt(x)) #