3.6 Question 215

[Eigenvalue]

For the following exercises, given y=f(u) and u=g(x), find [tex]\frac{dy}{dx}[/tex] by using Leibniz's notation for the chain rule: [tex]\frac{dy}{dx}[/tex]=[tex]\frac{dy}{du}[/tex]*[tex]\frac{du}{dx}[/tex]

y=6u³
u=7x-4

[Eigenvalue]

1. Find [tex]\frac{dy}{du}[/tex]
[tex]\frac{dy}{du}[/tex]=(6u³)'=6(3[tex]u^{3-1 }[/tex])=18u²
2. Find [tex]\frac{du}{dx}[/tex]
[tex]\frac{du}{dx}[/tex]=(7x-4)'=7
3. Find the product of [tex]\frac{dy}{du}[/tex] and [tex]\frac{dx}{du}[/tex]
[tex]\frac{dy}{dx}[/tex]=7*18u²=126u²
4.Substitute 7x-4 for u
[tex]\frac{dy}{dx}[/tex]=126(7x-4)²

Check:
1. Substitute 7x-4=u
y=6(7x-4)³
2. Differentiate both sides to find [tex]\frac{dy}{dx}[/tex]
[tex]\frac{dy}{dx}[/tex]=6(3[tex](7x-4)^{(3-1)}[/tex]*(7x-4)')
=6(3(7x-4)²*7)
=126(7x-4)²

Note: there is an additional method, that involves binomial expansion, to double check