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

##### 1 Answer

#### Answer:

# 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 "

# 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

# { ("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 = (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)) #