Question

differentiate the following function.
( y = \frac { 7 } { sqrt { x } + 4 } )
( \frac { d } { d x } left \frac { 7 } { sqrt { x } + 4 }
ight = )

Explanation:

Step1: Rewrite the function

Rewrite \(y = \frac{7}{\sqrt{x}+4}=7(\sqrt{x}+4)^{- 1}\), where \(\sqrt{x}=x^{\frac{1}{2}}\).

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 \(u = \sqrt{x}+4=x^{\frac{1}{2}}+4\), so \(y = 7u^{-1}\). First, find \(\frac{dy}{du}\): \(\frac{dy}{du}=-7u^{-2}=-\frac{7}{u^{2}}\). Then find \(\frac{du}{dx}\): \(\frac{du}{dx}=\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}}\).

Step3: Multiply the derivatives

By the chain rule \(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\). Substitute \(u = \sqrt{x}+4\) into \(\frac{dy}{du}\), we get \(\frac{dy}{dx}=-\frac{7}{(\sqrt{x}+4)^{2}}\cdot\frac{1}{2\sqrt{x}}\).

Step4: Simplify the expression

\(\frac{dy}{dx}=-\frac{7}{2\sqrt{x}(\sqrt{x}+4)^{2}}\)

Answer:

\(-\frac{7}{2\sqrt{x}(\sqrt{x}+4)^{2}}\)