Question

find the derivative of the function y = \sqrt{5 - 7x}.
\frac{dy}{dx}=\square

Explanation:

Step1: Rewrite the function

Rewrite $y = \sqrt{5 - 7x}$ as $y=(5 - 7x)^{\frac{1}{2}}$.

Step2: Apply the chain - rule

The chain - rule states that if $y = f(g(x))$, then $y^\prime=f^\prime(g(x))\cdot g^\prime(x)$. Let $u = 5 - 7x$, so $y = u^{\frac{1}{2}}$. First, find $\frac{dy}{du}$ and $\frac{du}{dx}$.
For $y = u^{\frac{1}{2}}$, $\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}$ by the power rule $\frac{d}{du}(u^n)=nu^{n - 1}$.
For $u = 5 - 7x$, $\frac{du}{dx}=-7$.

Step3: Calculate $\frac{dy}{dx}$

By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. Substitute $\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}$ and $\frac{du}{dx}=-7$ into the formula. Replace $u$ with $5 - 7x$:
$\frac{dy}{dx}=\frac{1}{2}(5 - 7x)^{-\frac{1}{2}}\cdot(-7)=-\frac{7}{2\sqrt{5 - 7x}}$.

Answer:

$-\frac{7}{2\sqrt{5 - 7x}}$