How do you differentiate #f(x)=csc(1/x^3) # using the chain rule?

1 Answer
Mar 23, 2017

Answer:

# f'(x) = (3csc(1/x^3)cot(1/x^3))/x^4 #

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 = csc(1/x^3) #, Then:

# { ("Let "u=1/x^3, => , (du)/dx=-3/x^4), ("Then "y=cscu, =>, dy/(du)=-cscucotu ) :}#

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

# \ \ \ \ \ dy/dx = (-cscucotu)(-3/x^4) #
# \ \ \ \ \ \ \ \ \ \ = (3cscucotu)/x^4 #
# \ \ \ \ \ \ \ \ \ \ = (3csc(1/x^3)cot(1/x^3))/x^4 #