find f(x).\nf(x)=(7 - 5x)^{13}\nf(x)=□

Explanation:

Step1: Identify the outer - inner functions

Let $u = 7 - 5x$, then $y = u^{13}$.

Step2: Differentiate the outer function

The derivative of $y$ with respect to $u$ is $\frac{dy}{du}=13u^{12}$ (using the power rule $\frac{d}{du}(u^n)=nu^{n - 1}$ with $n = 13$).

Step3: Differentiate the inner function

The derivative of $u$ with respect to $x$ is $\frac{du}{dx}=-5$ (since $\frac{d}{dx}(7 - 5x)=\frac{d}{dx}(7)-5\frac{d}{dx}(x)=0 - 5=-5$).

Step4: Apply the chain - rule

By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. Substitute $\frac{dy}{du}=13u^{12}$ and $\frac{du}{dx}=-5$ into the chain - rule formula. Replace $u$ with $7 - 5x$. So $\frac{dy}{dx}=13(7 - 5x)^{12}\cdot(-5)$.

Answer:

$-65(7 - 5x)^{12}$