Question

find f(x). f(x)=(5x^3 + 7)^6 f(x)=□

Explanation:

Step1: Identify the outer - inner functions

Let $u = 5x^{3}+7$, then $y = u^{6}$.

Step2: Differentiate the outer function

The derivative of $y$ with respect to $u$ is $\frac{dy}{du}=6u^{5}$ (using the power rule $\frac{d}{du}(u^{n})=nu^{n - 1}$ with $n = 6$).

Step3: Differentiate the inner function

The derivative of $u$ with respect to $x$ is $\frac{du}{dx}=15x^{2}$ (using the power rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$ for $a = 5,n = 3$ and $\frac{d}{dx}(c)=0$ for $c = 7$).

Step4: Apply the chain - rule

By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. Substitute $u = 5x^{3}+7$, $\frac{dy}{du}=6u^{5}$, and $\frac{du}{dx}=15x^{2}$ into the chain - rule formula. We get $\frac{dy}{dx}=6(5x^{3}+7)^{5}\cdot15x^{2}=90x^{2}(5x^{3}+7)^{5}$.

Answer:

$90x^{2}(5x^{3}+7)^{5}$