find the derivative of the function. y = 7tan^(-1)(x - √(1 + x^2)) y =

Explanation:

Step1: Recall chain - rule

The derivative of $y = 7\tan^{- 1}(u)$ with respect to $x$ is $y'=\frac{7}{1 + u^{2}}\cdot u'$, where $u=x-\sqrt{1 + x^{2}}$.

Step2: Find the derivative of $u$

Let $u=x-\sqrt{1 + x^{2}}$. The derivative of $x$ with respect to $x$ is $1$. For $v = \sqrt{1 + x^{2}}=(1 + x^{2})^{\frac{1}{2}}$, by the chain - rule, $v'=\frac{1}{2}(1 + x^{2})^{-\frac{1}{2}}\cdot2x=\frac{x}{\sqrt{1 + x^{2}}}$. So $u'=1-\frac{x}{\sqrt{1 + x^{2}}}=\frac{\sqrt{1 + x^{2}}-x}{\sqrt{1 + x^{2}}}$.

Step3: Substitute $u$ and $u'$ into the chain - rule formula

$y'=\frac{7}{1+(x - \sqrt{1 + x^{2}})^{2}}\cdot\frac{\sqrt{1 + x^{2}}-x}{\sqrt{1 + x^{2}}}$. First, expand $1+(x - \sqrt{1 + x^{2}})^{2}=1+x^{2}-2x\sqrt{1 + x^{2}}+1 + x^{2}=2 + 2x^{2}-2x\sqrt{1 + x^{2}}=2(1 + x^{2}-x\sqrt{1 + x^{2}})$. Now, $y'=\frac{7(\sqrt{1 + x^{2}}-x)}{\sqrt{1 + x^{2}}\cdot2(1 + x^{2}-x\sqrt{1 + x^{2}})}$. Rationalize the denominator: [ \begin{align*} y'&=\frac{7(\sqrt{1 + x^{2}}-x)}{\sqrt{1 + x^{2}}\cdot2(1 + x^{2}-x\sqrt{1 + x^{2}})}\cdot\frac{\sqrt{1 + x^{2}}+x}{\sqrt{1 + x^{2}}+x}\ &=\frac{7((\sqrt{1 + x^{2}}-x)(\sqrt{1 + x^{2}}+x))}{2(1 + x^{2}-x\sqrt{1 + x^{2}})(1 + x^{2})}\ &=\frac{7(1 + x^{2}-x^{2})}{2(1 + x^{2}-x\sqrt{1 + x^{2}})(1 + x^{2})}\ &=\frac{7}{2(1 + x^{2}-x\sqrt{1 + x^{2}})(1 + x^{2})}\cdot(1)\ &=\frac{7}{\sqrt{1 + x^{2}}} \end{align*} ]

Answer:

$\frac{7}{\sqrt{1 + x^{2}}}$