Derivative of 1/(1+x) by First Principle

The derivative of 1/(1+x) is equal to -1/(1+x)². In this post, we will find the derivative of 1 divided by 1+x using the limit definition, that is, by the first principle.

The first principle of derivatives says that the derivative of a function f(x) is given by the following limit:

$\dfrac{d}{dx}(f(x))$ $=\lim\limits_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$ …(I)

Derivative of 1/(1+x) using Limit Definition

To find the derivative of 1/1+x using definition, we will put f(x)=$\frac{1}{1+x}$ in the above Equation (I).

So we have

f(x+h) = $\dfrac{1}{1+x+h}$

Therefore, f(x+h)-f(x) = $\dfrac{1}{1+x+h}-\dfrac{1}{1+x}$

= $\dfrac{1+x-1-x-h}{(1+x+h)(1+x)}$

= $\dfrac{-h}{(1+x+h)(1+x)}$

So using (I), the derivative of 1/1+x is equal to

$\dfrac{d}{dx}(\dfrac{1}{1+x})$ = $\lim\limits_{h \to 0} \dfrac{\frac{-h}{(1+x+h)(1+x)}}{h}$

= $\lim\limits_{h \to 0} \Big[\dfrac{-h}{(1+x+h)(1+x)} \times \dfrac{1}{h} \Big]$

= $\lim\limits_{h \to 0} \dfrac{-1}{(1+x+h)(1+x)}$

= $\dfrac{-1}{(1+x+0)(1+x)}$

= $\dfrac{-1}{(1+x)^2}$

Thus, the derivative of 1/1+x using definition is equal to -1/(1+x)². This is proved by the first principle of derivatives.

Question: What is the derivative of 1/(1+x) at x=1?

Answer:

From above, we obtain that

$\Big[ \dfrac{d}{dx}(\dfrac{1}{1+x})\Big]_{x=1}$ = $\Big[ \dfrac{-1}{(1+x)^2}\Big]_{x=1}$

= $\dfrac{-1}{(1+1)^2}$

= $\dfrac{-1}{2^2}$

= $\dfrac{-1}{4}$

So the derivative of 1/(1+x) at x=1 is equal to -1/4.

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