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

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})