work: section 3.4 the chain\nquestion 5, 3.4.21\nfind f(x).\nf(x)=(3x^6 + 1)^5\nf(x)=□

Explanation:

Step1: Identify outer - inner functions

Let $u = 3x^{6}+1$, so $y = u^{5}$.

Step2: Differentiate outer function

The derivative of $y$ with respect to $u$ is $\frac{dy}{du}=5u^{4}$ (using the power rule $\frac{d}{du}(u^{n})=nu^{n - 1}$ with $n = 5$).

Step3: Differentiate inner function

The derivative of $u$ with respect to $x$ is $\frac{du}{dx}=18x^{5}$ (using the power rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$ for $a = 3$ and $n = 6$).

Step4: Apply chain - rule

By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. Substitute $u = 3x^{6}+1$, $\frac{dy}{du}=5u^{4}$, and $\frac{du}{dx}=18x^{5}$ into the chain - rule formula. We get $\frac{dy}{dx}=5(3x^{6}+1)^{4}\cdot18x^{5}$.

Step5: Simplify

$\frac{dy}{dx}=90x^{5}(3x^{6}+1)^{4}$.

Answer:

$90x^{5}(3x^{6}+1)^{4}$