Question

Find the derivative of
`tan^(-1) x` by 1st principle of derivative.

Answer 1

To find the derivative of
`\tan^(-1) x` using the first principle of derivative, we need to use the definition of the derivative:

f'(x) = lim(h->0) [(f(x + h) - f(x)) / h]

where f(x) =
`\tan^(-1) x`.

Substituting f(x) into the definition of the derivative, we get:

f'(x) = lim(h->0) [(
`\tan^(-1)`(x + h) -
(a + b) =
`\tan^(-1)`a +
`\tan^(-1)`b -
`\pi`/2

Using this formula, we can rewrite the numerator of the expression above as:

(
`\tan^(-1)`(x + h) -
`\tan^(-1)/tex) = [tex]\tan^(-1)`((x + h) / (1 + (x + h)^2)) -
`\tan^(-1)`(x / (1 + x^2))

Now, substituting this expression back into the definition of the derivative, we get:

f'(x) = lim(h->0) [
`\tan^(-1)`((x + h) / (1 + (x + h)^2)) -
`\tan^(-1)`(x / (1 + x^2))] / h

We can simplify this expression using algebra and trigonometry, and we get:

f'(x) = lim(h->0) [h / (1 + x^2 + hx + h^2 + x^2h + xh^2)] / h

f'(x) = lim(h->0) 1 / (1 + x^2 + hx + h^2 + x^2h + xh^2)

Now we can simplify this expression by dropping the terms that contain h^2 or higher powers of h, since they will approach zero faster than h as h approaches zero. We also drop the term containing x^2h, since it is a second-order term and will also approach zero faster than h. This leaves us with:

f'(x) = lim(h->0) 1 / (1 + x^2 + hx)

Now we can evaluate the limit as h approaches zero:

f'(x) = 1 / (1 + x^2)

Therefore, the derivative of
`\tan^(-1) x` by first principle of derivative is:

`(d)/(dx)`
`\tan^(-1) x` = 1 / (1 + x^2)

Answer 2

`\frac{\text{d}}{\text{d}x} \tan^(-1)x=(1)/(1+x^2)`

Explanation:

`\boxed{\begin{minipage}{6.5 cm}\underline{Trigonometric Identity}\\\\$\tan^(-1)(A)-\tan^(-1)(B) \equiv \tan^(-1)\left((A-B)/(1+AB)\right)$\\\end{minipage}}`

`\boxed{\begin{minipage}{3 cm}$\displaystyle \lim_(h \to 0) \left[(\tan^(-1) \theta)/(\theta) \right]=1$\\\end{minipage}}`

`\boxed{\begin{minipage}{5.6 cm}\underline{Differentiating from First Principles}\\\\$\text{f}\:'(x)=\displaystyle \lim_(h \to 0) \left[\frac{\text{f}(x+h)-\text{f}(x)}{(x+h)-x}\right]$\\\end{minipage}}`

Given function:

`\text{f}(x)=\tan^(-1)x`

`\implies \text{f}(x+h)=\tan^(-1)(x+h)`

Differentiating from first principles:

`\begin{aligned}\text{f}\:'(x) & =\displaystyle \lim_(h \to 0) \left[\frac{\text{f}(x+h)-\text{f}(x)}{(x+h)-x}\right]\\\\& =\lim_(h \to 0) \left[(\tan^(-1)(x+h)-\tan^(-1)x)/((x+h)-x)\right]\end{aligned}`

Using the trigonometric identity to rewrite the numerator:

`\begin{aligned}& =\lim_(h \to 0) \left[(\tan^(-1)\left((x+h-x)/(1+x(x+h))\right))/((x+h)-x)\right]\\\\& =\lim_(h \to 0) \left[(\tan^(-1)\left((h)/(1+x^2+xh))\right))/(h)\right]\end{aligned}`

`\textsf{Multiply the denominator by }(1+x^2+xh)/(1+x^2+xh):`

`= \displaystyle \lim_(h \to 0) \left[(\tan^(-1)\left((h)/(1+x^2+xh))\right))/((h(1+x^2+xh))/((1+x^2+xh)))\right]`

Separate:

`= \displaystyle \lim_(h \to 0) \left[(\tan^(-1)\left((h)/(1+x^2+xh))\right))/((h)/((1+x^2+xh))) \right] \cdot \displaystyle \lim_(h \to 0) \left[(1)/(1+x^2+xh)\right]`

`\textsf{Use }\displaystyle \lim_(h \to 0) \left[(\tan^(-1) \theta)/(\theta) \right]=1:`

`= 1 \cdot \displaystyle \lim_(h \to 0) \left[(1)/(1+x^2+xh)\right]`

As h gets close to zero:

`= 1 \cdot \left[(1)/(1+x^2)\right]`

Simplify:

`=(1)/(1+x^2)`