How do you use the chain rule to differentiate y=1/(x^4+x)^2?
2 Answers
Explanation:
We're asked to find the derivative
(dy)/(dx) [y = 1/((x^4 + x)^2)]
We can first use the quotient rule, which is
d/(dx) [u/v] = (v(du)/(dx) - u(dv)/(dx))/(v^2)
where
-
u = 1 -
v = (x^4 + x)^2 :
y'(x) = ((x^4+x)^2d/(dx)[1] - 1d/(dx) [(x^4+x)^2])/((x^4+x)^4)
The derivative of
y'(x) = (-1d/(dx) [(x^4+x)^2])/((x^4+x)^4)
Now, we'll use the chain rule:
d/(dx) [(x^4+x)^2] = d/(du) [u^2] (du)/(dx)
where
-
u = x^4 + x -
d/(du) [u^2] = 2u (from power rule):
y'(x) = (-2(x^4+x)d/(dx)[x^4+x])/((x^4+x)^4)
The derivative of
y'(x) = (-2(x^4+x)(4x^3 + 1))/((x^4+x)^4)
Which simplifies to
color(blue)(ulbar(|stackrel(" ")(" "y'(x) = (-2(4x^3+1))/((x^4+x)^3)" ")|)
Explanation:
Alternatively, we could rewrite as
y = (x^4 + x)^-2
Letting
y' = -2u^-3 * 4x^3 + 1
y' = -2(x^4 + x)^(-3) * 4x^3 + 1
y' = -(8x^3 + 2)/(x^4 + x)^3
Hopefully this helps!