Question

find the derivative of the function ( y = sqrt{4 - 5x} ).

( \frac{dy}{dx} = square )

Explanation:

Step1: Rewrite the function

Rewrite \(y = \sqrt{4-5x}\) as \(y=(4 - 5x)^{\frac{1}{2}}\).

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 \(u = 4-5x\), so \(y = u^{\frac{1}{2}}\). First, find \(\frac{dy}{du}\): \(\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}\). Then find \(\frac{du}{dx}\): \(\frac{du}{dx}=- 5\).

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 \(\frac{du}{dx}=-5\) into the formula. Since \(u = 4-5x\), we have \(\frac{dy}{dx}=\frac{1}{2}(4 - 5x)^{-\frac{1}{2}}\cdot(-5)\).
Simplify the expression: \(\frac{dy}{dx}=-\frac{5}{2\sqrt{4 - 5x}}\).

Answer:

\(-\frac{5}{2\sqrt{4 - 5x}}\)