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

Explanation:

Step1: Recall chain - rule

The chain - rule states that if $y = f(g(x))$, then $y'=f'(g(x))\cdot g'(x)$. Here $f(u) = 7\tan^{- 1}(u)$ and $u=x-\sqrt{1 + x^{2}}$. First, find $f'(u)$. The derivative of $\tan^{-1}(u)$ is $\frac{1}{1 + u^{2}}$, so $f'(u)=\frac{7}{1 + u^{2}}$.

Step2: Find the derivative of $u$

Let $u=x-\sqrt{1 + x^{2}}$. The derivative of $x$ with respect to $x$ is $1$. For $v=\sqrt{1 + x^{2}}=(1 + x^{2})^{\frac{1}{2}}$, using the chain - rule: if $v=(1 + x^{2})^{\frac{1}{2}}$, let $t = 1+x^{2}$, then $v=t^{\frac{1}{2}}$. The derivative of $v$ with respect to $t$ is $\frac{1}{2}t^{-\frac{1}{2}}$ and the derivative of $t$ with respect to $x$ is $2x$. So $v'=\frac{1}{2}(1 + x^{2})^{-\frac{1}{2}}\cdot2x=\frac{x}{\sqrt{1 + x^{2}}}$. Then $u'=1-\frac{x}{\sqrt{1 + x^{2}}}=\frac{\sqrt{1 + x^{2}}-x}{\sqrt{1 + x^{2}}}$.

Step3: Apply the chain - rule

Substitute $u=x-\sqrt{1 + x^{2}}$ into $f'(u)$ and multiply by $u'$. [ \begin{align*} y'&=\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}})}\ &=\frac{7}{2\sqrt{1 + x^{2}}} \end{align*} ]

Answer:

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