How do you differentiate #(5x^2+pi^2)^2#?

1 Answer
Nov 3, 2016

# dy/dx=20x(5x^2+pi^2) #

Explanation:

If 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) #

Eg If # y=(5x^2+pi^2)^2 #, Then:

# { ("Let "u=5x^2+pi^2, => , (du)/dx=10x), ("Then "y=u^2, =>, dy/(du)=2u ) :}#

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

# dy/dx=(2u)(10x) #
# :. dy/dx=2(5x^2+pi^2)(10x) #
# :. dy/dx=20x(5x^2+pi^2) #