Question

write the following function in the form y = f(u) and u = g(x). then find $\frac{dy}{dx}$ as a function of x.
$y = (\frac{x^{2}}{6}+3x - \frac{4}{x})^{6}$
c. y = f(u)=u and u = g(x)=$\frac{x^{2}}{6}+3x$
d. y = f(u)=u and u = g(x)=$\frac{x^{2}}{6}+3x - \frac{4}{x}$
find $\frac{dy}{dx}$ as a function of x.
$\frac{dy}{dx}=square$

Explanation:

Step1: Identify f(u) and g(x)

Let $u = g(x)=\frac{x^{2}}{6}+3x - \frac{4}{x}$, and $y = f(u)=u^{6}$.

Step2: Find $\frac{dy}{du}$

Using the power - rule, if $y = u^{6}$, then $\frac{dy}{du}=6u^{5}$.

Step3: Find $\frac{du}{dx}$

Differentiate $u=\frac{x^{2}}{6}+3x - \frac{4}{x}=\frac{1}{6}x^{2}+3x - 4x^{-1}$ with respect to $x$.
$\frac{du}{dx}=\frac{1}{6}\times2x + 3+4x^{-2}=\frac{x}{3}+3+\frac{4}{x^{2}}$.

Step4: Use the chain - rule

The chain - rule states that $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$.
Substitute $u=\frac{x^{2}}{6}+3x - \frac{4}{x}$ and $\frac{dy}{du}=6u^{5}$, $\frac{du}{dx}=\frac{x}{3}+3+\frac{4}{x^{2}}$ into the chain - rule formula:
$\frac{dy}{dx}=6(\frac{x^{2}}{6}+3x - \frac{4}{x})^{5}(\frac{x}{3}+3+\frac{4}{x^{2}})$.

Answer:

$6(\frac{x^{2}}{6}+3x - \frac{4}{x})^{5}(\frac{x}{3}+3+\frac{4}{x^{2}})$