How do you use the chain rule to differentiate #y=5/(2x^3+3x)#?

1 Answer
Steve M
Dec 6, 2016

# dy/dx = -5((6x^2+3))/(2x^3+3x)^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) #

So with # y = 5/(2x^3+3x) #, Then:

# { ("Let "u=2x^3+3x, => , (du)/dx=6x^2+3), ("Then "y=5/u=u^-1, =>, dy/(du)=-u^-2=-5/u^2 ) :}#

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

# dy/dx = (-5/u^2)(6x^2+3) #
# :. dy/dx = -5(6x^2+3)/(2x^3+3x)^2 #

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