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Chapter 6: Problem 5

Find the derivative of \(y=f(x)=\tan ^{-1} x\), using first principle of differentiation.

Short Answer

Expert verified
The derivative of \(y=f(x)=tan^{-1}x\) using first principle of differentiation is \(f'(x)=1/(1+x^2)\).

Step by step solution

01

Write down the original function

Given the original function \(y=f(x)=tan^{-1}x\).
02

Apply the formula for first principle of differentiation

The formula for first principle of differentiation is given by \(\lim_{h\to0} \frac{f(x+h)-f(x)}{h}\). Replace \(f(x)\) and \(f(x+h)\) with \(tan^{-1}x\) and \(tan^{-1}(x + h)\) respectively, which gives \(f'(x)=\lim_{h\to0} \frac{tan^{-1}(x+h)-tan^{-1}x}{h}\).
03

Use the identity of \(tan^{-1} a - tan^{-1} b = tan^{-1}((a-b)/(1+ab))\)

By the identity, the preceding step simplifies to \(f'(x)=\lim_{h\to0} \frac{tan^{-1}((x+h - x)/ (1 + x(x + h)))}{h}\), which simplifies further to \(f'(x)=\lim_{h\to0} \frac{tan^{-1}(h/ (1 + x^2 + hx))}{h}\). In the limit of \(h\) approaching 0, the term \(hx\) can be ignored. Hence we get \(f'(x)=\lim_{h\to0} \frac{tan^{-1}(h/ (1 + x^2))}{h}\).
04

Take the limit as h tends to 0

As \(h\to 0\), we know that \(tan^{-1}a/a \to 1\). So, \(f'(x)=\lim_{h\to0} \frac{h/ (1 + x^2)}{h}\). After cancelling out \(h\), we get \(f'(x)=1/(1+x^2)\).

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