Question

1. (6 points) given f(x)=(2x - 2)^4

a) find dy = f(x)·dx
b) evaluate dy when x = 4 and dx = 0.02

Explanation:

Step1: Apply chain - rule

Let $u = 2x - 2$, then $y = u^{4}$. The chain - rule states that $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. First, find $\frac{dy}{du}$ and $\frac{du}{dx}$. $\frac{dy}{du}=4u^{3}$ and $\frac{du}{dx}=2$. Substituting $u = 2x - 2$ back in, we get $f^{\prime}(x)=\frac{dy}{dx}=4(2x - 2)^{3}\cdot2=8(2x - 2)^{3}$. So, $dy = 8(2x - 2)^{3}dx$.

Step2: Substitute values

We know that $x = 4$ and $dx=0.02$. Substitute $x = 4$ into $f^{\prime}(x)$: $f^{\prime}(4)=8(2\times4 - 2)^{3}=8\times6^{3}=8\times216 = 1728$. Then, since $dy = f^{\prime}(x)dx$, when $x = 4$ and $dx = 0.02$, $dy=1728\times0.02 = 34.56$.

Answer:

a) $dy = 8(2x - 2)^{3}dx$
b) $34.56$