How to find the derivative of 1/x. In this article, we will find the derivative of 1 divided by x using the power rule, product rule, and the definition of derivatives.
What is the Derivative of 1/x?
At first, we will evaluate the derivative of 1/x by the power rule of derivatives. We need to follow the below steps.
Step 1: First, we will express 1/x as a power of x using the rule of indices. So we have
$1/x=x^{-1}$
Step 2: Now, we will apply the power rule of derivatives: $\frac{d}{dx}(x^n)=nx^{n-1}$. Thus we get that
$\frac{d}{dx}(1/x)=\frac{d}{dx}(x^{-1})=-1 \cdot x^{-1-1}$
Step 3: Simplifying the above expression, we obtain that
$\dfrac{d}{dx}(\dfrac{1}{x})=-1 \cdot x^{-2}$
$\Rightarrow \dfrac{d}{dx}(\dfrac{1}{x})=-1 \cdot \dfrac{1}{x^2}$
$\Rightarrow \dfrac{d}{dx}(\dfrac{1}{x})=\dfrac{-1}{x^2}$
So the derivative of $1/x$ by the power rule is $-1/x^2$. In other words,
\[\dfrac{d}{dx}(\frac{1}{x})=-\dfrac{1}{x^2}.\]
Also Read:
Derivative of esin x : The derivative of esin x is cos x esin x.
Integration of mod x : The integration of mod x is -x|x|/2+c
Now, we will find the derivative of 1/x by the first principle.
Derivative of 1/x from first principle
Let $f(x)=\dfrac{1}{x}.$ Applying the first principle of derivatives, we get that
$\dfrac{d}{dx}(f(x))$ $=\lim\limits_{h \to 0}\dfrac{f(x+h)-f(x)}{h}$
From the above definition of derivatives, the derivative of 1/x by first principle is equal to
$\dfrac{d}{dx}(\dfrac{1}{x})$ $= \lim\limits_{h \to 0} \dfrac{\frac{1}{x+h}-\frac{1}{x}}{h}$
$=\lim\limits_{h \to 0}\dfrac{\frac{x-x-h}{x(x+h)}}{h}$
$=\lim\limits_{h \to 0}\dfrac{-h}{hx(x+h)}$
$=-\lim\limits_{h \to 0}\dfrac{1}{x(x+h)}$
$=-\dfrac{1}{x(x+0)}$
$=-\dfrac{1}{x^2}$
This shows that the formula of the derivative of 1/x is -1/x2. This is obtained by the first principle of derivatives.
We know that the product rule of derivatives is $\frac{d}{dx}(fg)=f \frac{dg}{dx}+ g \frac{df}{dx}$. Using this rule, we will now find the derivative of 1/x.
Derivative of 1/x by product rule
Let us put z=1/x. This implies that
zx=1
Differentiating with respect to x, we get that
$\dfrac{d}{dx}(zx)=\dfrac{d}{dx}(1)$
$\Rightarrow z\dfrac{d}{dx}(x)+x\dfrac{d}{dx}(z)=0$ (by the product rule of derivatives)
$\Rightarrow z\cdot 1+x\dfrac{dz}{dx}=0$
$\Rightarrow x\dfrac{dz}{dx}=-z$
$\Rightarrow \dfrac{dz}{dx}=-\dfrac{z}{x}$
$\Rightarrow \dfrac{dz}{dx}=-\dfrac{1}{x^2}$ as z=1/x
So we have obtained the derivative of 1/x by the product rule which is -1/x2.
Also Read:
Derivative of root x: The derivative of √x is 1/2√x
Derivative of cube root of x: The derivative of the cube root of x is 1/(3x^{2/3})
Derivative of sin inverse x: The derivative of sin-1 x is 1/√(1-x2)
Derivative of sin 3x: The derivative of sin 3x is 3cos 3x
Application of Derivative of 1/x
As an application of the derivative of 1/x, we will now find the derivative of 1/log x. We will use the chain rule of derivatives: $\frac{du}{dx}=\frac{du}{dz} \cdot \frac{dz}{dx}$
Question: What is the derivative of $\frac{1}{\log x}$
Solution:
Let $z=\log x$. So we have $\frac{dz}{dx}=\frac{1}{x}$
Now, $\dfrac{d}{dx}(\dfrac{1}{\log x})$
$=\dfrac{d}{dx}(\dfrac{1}{z})$
$=\dfrac{d}{dz}(\dfrac{1}{z}) \cdot \dfrac{dz}{dx}$ (by the chain rule)
$=-\dfrac{1}{z^2} \cdot \dfrac{1}{x}$ (by the above formula of the derivative of 1/x)
$=-\dfrac{1}{x(\log x)^2}$ as z=log x.
