How do you use the chain rule to differentiate y=(x^3+4)^5/(3x^4-2)?

1 Answer
Sep 28, 2016

color(blue)(y'=((x^3+4)^4(33x^6-48x^3-30x^2))/(3x^4-2)^2)

Explanation:

y is a quotient in the form of color(blue)(y=(u(x))/(v(x)))
The deferentiation of the quotient is as follows :

color(blue)(y'=((u(x))'v(x)-(v(x))'u(x))/(v(x))^2)
Let us find (u(x))' and (v(x))'

color(green)((u(x))'=?)
u(x) is a composite of two functions f(x) and g(x) where:
f(x)=x^5 and g(x)=x^3+4

We have to use chain rule to find color(green)((u(x))')

u(x)=f(g(x)) then
color(green)((u(x))'=f'(g(x))*g'(x))

f'(x)=5x^4 then
f'(g(x))=5(g(x))^4
color(green)(f'(g(x))=5(x^3+4)^4)

color(green)((g(x))'=3x^2)
So,(u(x))'=5(x^3+4)^4*3x^2
color(green)((u(x))'=15x^2(x^3+4)^4)

color(red)((v(x))'=?)

v(x)=3x^4-2
color(red)((v(x))'=12x^3)

Now,let us substitute color(green)((u(x))' and color(red)((v(x))' in color(blue)y'

color(blue)(y'=((u(x))'v(x)-(v(x))'u(x))/(v(x))^2)

y'=(color(green)(15x^2(x^3+4)^4)*(3x^4-2)-color(red)(12x^3)(x^3+4)^5)/(3x^4-2)^2

y'=((x^3+4)^4[15x^2(3x^4-2)-12x^3(x^3+4)])/(3x^4-2)^2

y'=((x^3+4)^4[45x^6-30x^2-12x^6-48x^3])/(3x^4-2)^2

y'=((x^3+4)^4(45x^6-12x^6-48x^3-30x^2))/(3x^4-2)^2
Therefore,

color(blue)(y'=((x^3+4)^4(33x^6-48x^3-30x^2))/(3x^4-2)^2)