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

Explanation:

Step1: Apply the chain rule

The chain rule states that if (y = 7\tan^{-1}(u)) where (u=x-\sqrt{1 + x^{2}}), then (y^\prime=7\times\frac{1}{1 + u^{2}}\times u^\prime).

Step2: Find the derivative of (u)

First, find (u^\prime). The derivative of (x) is (1). For (v = \sqrt{1 + x^{2}}=(1 + x^{2})^{\frac{1}{2}}), using the chain rule: (v^\prime=\frac{1}{2}(1 + x^{2})^{-\frac{1}{2}}\times2x=\frac{x}{\sqrt{1 + x^{2}}}). So (u^\prime=1-\frac{x}{\sqrt{1 + x^{2}}}=\frac{\sqrt{1 + x^{2}}-x}{\sqrt{1 + x^{2}}}).

Step3: Substitute (u) and (u^\prime) into (y^\prime)

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

Answer:

(\frac{7}{2(1 + x^{2})})