Derivative of 1/x^2 by First Principle

The derivative of 1 by x square is equal to -2/x³. In this blog post, we will learn how to find the differentiation of 1/x².

What is the Derivative of 1/x²?

Let us now find the derivative of 1/x^2 by the power rule of derivatives. The power rule says that

$\dfrac{d}{dx}(x^n)=nx^{n-1}$

Step 1: First,we write $1/x^2$ as follows:

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

Step 2: Differentiating with respect to x, we obtain that

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

Step 3: Apply the above power rule with $n=-2$. Thus, we have

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

$=-2x^{-2-1}$ $=-2x^{-3}$

$=-\dfrac{2}{x^3}$.

So the derivative of 1/x^2 is equal to -2/x^3.

More Derivatives:

Derivative of 1/lnx	Derivative of 1/2x
Derivative of √x	Derivative of 2x

Derivative of 1/x² by First Principle

Let $f(x)=1/x^2$. Then the derivative of 1/x2 by the first principle of derivatives will be equal to

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

∴ $\dfrac{d}{dx}(1/x^2)$ $=\lim\limits_{h \to 0} \dfrac{\dfrac{1}{(x+h)^2}-\dfrac{1}{x^2}}{h}$

$=\lim\limits_{h \to 0} \dfrac{x^2-(x+h)^2}{hx^2(x+h)^2}$

$=\lim\limits_{h \to 0} \dfrac{x^2-x^2-2xh-h^2}{hx^2(x+h)^2}$ as we know that $(a+b)^2$ $=a^2+2ab+b^2$

$=\lim\limits_{h \to 0} \dfrac{-2xh-h^2}{hx^2(x+h)^2}$

$=\lim\limits_{h \to 0} \dfrac{-h(2x+h)}{hx^2(x+h)^2}$

$=-\lim\limits_{h \to 0} \dfrac{2x+h}{x^2(x+h)^2}$

$=-\dfrac{2x+0}{x^2(x+0)^2}$ $=-\dfrac{2x}{x^4}$

$=-\dfrac{2}{x^3}$

Thus, the differentiation/ derivative of 1/x^2 by first principle is -2/x³.

Also Read:

Derivative of 1/√x	Derivative of (√x + 1/√x)
Derivative of x √x	Derivative of 1/(1+x^2)

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