find the derivative of the function.
$y = 7\tan^{-1}(x - \sqrt{1 + x^{2}})$

Question

Accepted Answer

# Explanation:
## Step1: Use the chain rule
Let $u = x-\sqrt{1 + x^{2}}$, then $y = 7\tan^{- 1}(u)$. The derivative of $y$ with respect to $u$ is $\frac{dy}{du}=\frac{7}{1 + u^{2}}$.
## Step2: Find the derivative of $u$
The derivative of $x$ is $1$. For $v=\sqrt{1 + x^{2}}=(1 + x^{2})^{\frac{1}{2}}$, using the chain rule $\frac{dv}{dx}=\frac{1}{2}(1 + x^{2})^{-\frac{1}{2}}\cdot2x=\frac{x}{\sqrt{1 + x^{2}}}$. So $\frac{du}{dx}=1-\frac{x}{\sqrt{1 + x^{2}}}=\frac{\sqrt{1 + x^{2}}-x}{\sqrt{1 + x^{2}}}$.
## Step3: Use the chain rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$
Substitute $u = x-\sqrt{1 + x^{2}}$ and $\frac{du}{dx}=\frac{\sqrt{1 + x^{2}}-x}{\sqrt{1 + x^{2}}}$ into $\frac{dy}{dx}$:
$$
\begin{align*}
\frac{dy}{dx}&=\frac{7}{1+(x - \sqrt{1 + x^{2}})^{2}}\cdot\frac{\sqrt{1 + x^{2}}-x}{\sqrt{1 + x^{2}}}\
&=\frac{7(\sqrt{1 + x^{2}}-x)}{\sqrt{1 + x^{2}}(1+x^{2}-2x\sqrt{1 + x^{2}}+1 + x^{2})}\
&=\frac{7(\sqrt{1 + x^{2}}-x)}{\sqrt{1 + x^{2}}(2 + 2x^{2}-2x\sqrt{1 + x^{2}})}\
&=\frac{7(\sqrt{1 + x^{2}}-x)}{2\sqrt{1 + x^{2}}(1 + x^{2}-x\sqrt{1 + x^{2}})}\
\end{align*}
$$
Multiply numerator and denominator by $(1 + x^{2}+x\sqrt{1 + x^{2}})$:
$$
\begin{align*}
\frac{dy}{dx}&=\frac{7(\sqrt{1 + x^{2}}-x)(1 + x^{2}+x\sqrt{1 + x^{2}})}{2\sqrt{1 + x^{2}}((1 + x^{2})^{2}-x^{2}(1 + x^{2}))}\
&=\frac{7(\sqrt{1 + x^{2}}(1 + x^{2})+x(1 + x^{2})-x(1 + x^{2})-x^{2}\sqrt{1 + x^{2}})}{2\sqrt{1 + x^{2}}(1 + x^{2})(1 + x^{2}-x^{2})}\
&=\frac{7(1 + x^{2})\sqrt{1 + x^{2}}-x^{2}\sqrt{1 + x^{2}}}{2\sqrt{1 + x^{2}}(1 + x^{2})}\
&=\frac{7\sqrt{1 + x^{2}}}{2\sqrt{1 + x^{2}}}\
&=\frac{7}{2}
\end{align*}
$$

# Answer:
$\frac{7}{2}$

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QUESTION IMAGE

Question

Explanation:

Step1: Use the chain rule

Step2: Find the derivative of \(u\)

Step3: Use the chain rule \(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\)

Answer: