3.6 Question 214

[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=3u-6
u=2x²

[Eigenvalue]

1. Find [tex]\frac{dy}{du}[/tex] by differentiating y=3u-6:
[tex]\frac{dy}{du}[/tex]=(3u-6)'=3
2. Find [tex]\frac{du}{dx}[/tex] by differentiating u=2x²
[tex]\frac{du}{dx}[/tex]=(2x²)'=2(2[tex]x^{(2-1)}[/tex])=4x
3. Find the product of [tex]\frac{dy}{du}[/tex] and [tex]\frac{du}{dx}[/tex]
[tex]\frac{dy}{du}[/tex]*[tex]\frac{du}{dx}[/tex]=3*4x=12x

Check:
1. Double check by substituting 2x² for u:
y=3(2x²)-6=6x²-6
2. Differentiate both sides to find [tex]\frac{dy}{dx}[/tex]
[tex]\frac{dy}{dx}[/tex]=(6x²-6)'=6(2[tex]x^{(2-1) }[/tex])-0=12x
12x=12x