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

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}})