Question

find the derivative of the function.

( y = 7 \tan ^ { - 1 } left( x - sqrt { 1 + x ^ { 2 } }
ight) )

( y ^ { prime } = )

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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 \(y^\prime\).

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Answer:

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