Derivative of 1/(1+x): Proof by Chain, Quotient Rule

The derivative of 1/(1+x) is equal to -1/(1+x)². The function 1/(1+x) is the reciprocal of 1+x. In this post, we will learn how to differentiate 1 divided by 1+x. Its derivative is denoted by $\dfrac{d}{dx} \Big(\dfrac{1}{1+x} \Big)$ and it is equal to

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

We will use the chain rule and the quotient rule to show it.

Derivative of 1/(1+x) by Chain Rule

Let us now find the derivative of 1/(1+x) by the chain rule. We put

z=1+x.

So $\dfrac{dz}{dx}=1$.

Then by the chain rule of derivatives, we have

$\dfrac{d}{dx} \Big(\dfrac{1}{1+x} \Big)$ = $\dfrac{d}{dz} \Big(\dfrac{1}{z} \Big) \times \dfrac{dz}{dx}$

= $\dfrac{d}{dz} (z^{-1}) \times 1$

= -1z^-1-1 by the power rule of derivatives d/dx(xⁿ)=nx^n-1

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

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

So the derivative of 1/(1+x) is -1/(1+x)² which is proved by the chain rule and the power rule of derivatives.

Derivative of 1/(1+x) by Quotient Rule

As $\frac{1}{1+x}$ is a quotient function, we use the quotient rule to find its derivative. The rule tells us that the derivative of u/v is equal to

$\dfrac{d}{dx}\Big( \dfrac{u}{v}\Big)$ = $\dfrac{v \frac{du}{dx}-u \frac{dv}{dx}}{v^2}$

where u and v are two functions of x.

Here we put u=1, v=1+x.

So the derivative of 1/(1+x) by the quotient rule is equal to

$\dfrac{d}{dx} \Big(\dfrac{1}{1+x} \Big)$ = $\dfrac{(1+x)\cdot \frac{d}{dx}(1) – 1 \cdot \frac{d}{dx}(1+x)}{(1+x)^2}$

= $\dfrac{0- 1}{(1+x)^2}$ as $\frac{d}{dx}(1+x)=1$ and $\frac{d}{dx}(1)=0$

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

Thus, the derivative of 1/(1+x) is -1/(1+x)² which is proved by the quotient rule of derivatives.

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