find the derivative of the function.\ny = \\sqrt{\\frac{x}{x + 5}}\ny=\n\nresources\nread it watch it master…

Explanation:

Step1: Rewrite the function

First, rewrite $y = \sqrt{\frac{x}{x + 5}}$ as $y=\left(\frac{x}{x + 5}\right)^{\frac{1}{2}}$.

Step2: Apply the chain - rule

The chain - rule states that if $y = u^{\frac{1}{2}}$ and $u=\frac{x}{x + 5}$, then $y^\prime=\frac{1}{2}u^{-\frac{1}{2}}\cdot u^\prime$. First, find $u^\prime$ using the quotient - rule. The quotient - rule for $u=\frac{f(x)}{g(x)}$ where $f(x)=x$ and $g(x)=x + 5$ is $u^\prime=\frac{f^\prime(x)g(x)-f(x)g^\prime(x)}{g(x)^2}$. Here, $f^\prime(x)=1$ and $g^\prime(x)=1$, so $u^\prime=\frac{1\cdot(x + 5)-x\cdot1}{(x + 5)^2}=\frac{x + 5-x}{(x + 5)^2}=\frac{5}{(x + 5)^2}$. Then $y^\prime=\frac{1}{2}\left(\frac{x}{x + 5}\right)^{-\frac{1}{2}}\cdot\frac{5}{(x + 5)^2}=\frac{5}{2\sqrt{\frac{x}{x + 5}}(x + 5)^2}=\frac{5}{2\sqrt{x}(x + 5)^{\frac{3}{2}}}$.

For $y = e^{\tan(\theta)}$, apply the chain - rule.

Step1: Identify inner and outer functions

Let $u = \tan(\theta)$ and $y=e^{u}$.

Step2: Find derivatives of inner and outer functions

The derivative of $y$ with respect to $u$ is $\frac{dy}{du}=e^{u}$, and the derivative of $u$ with respect to $\theta$ is $\frac{du}{d\theta}=\sec^{2}(\theta)$.

Step3: Apply the chain - rule

By the chain - rule $\frac{dy}{d\theta}=\frac{dy}{du}\cdot\frac{du}{d\theta}=e^{\tan(\theta)}\sec^{2}(\theta)$.

Answer:

For $y = \sqrt{\frac{x}{x + 5}}$, $y^\prime=\frac{5}{2\sqrt{x}(x + 5)^{\frac{3}{2}}}$; for $y = e^{\tan(\theta)}$, $y^\prime=e^{\tan(\theta)}\sec^{2}(\theta)$