Question

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

Answer 1

`f'(x) = -(csc^2(sqrt(x)) * e^(cot sqrt(x))) / (2sqrt(x))`

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 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))`