# Derivative of 1/x^3: Formula, Proof by First Principle

The derivative of 1/x3 is equal to -3/x4. In this article, we will learn how to find the derivative of 1 divided by x3 using the power rule, product rule, and the definition of derivatives.

The derivative of 1/x3 can be expressed mathematically as d/dx(1/x3) or (1/x3)$’$. The derivative formula of 1 divided by x cube is given below:

d/dx(1/x3) = -3/x4 or (1/x3)$’$ = -3/x4.

To find the differentiation of 1 divided by x3, we will follow the below steps as described here:

Step1: Express 1/x3 as a power of x:

Note that 1/x3 is an algebraic function which can be expressed as x-3. That is,

1/x3 = x-3

Step2: Differentiate both sides w.r.t. x, we get that

d/dx(1/x3) = d/dx(x-3)

Step3: Apply the power rule of derivatives d/dx(xn) = nxn-1 with n=-3. So from the above we get that

d/dx(1/x3) = d/dx(x-3) = -3x-3-1

Simplifying, we deduce that

d/dx(1/x3) = -3x-4 = -3/x4.

Conclusion: The derivative of 1/x3 by power rule is -3/x4.

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

$\dfrac{d}{dx}(f(x))$ = limh→0 $\dfrac{f(x+h)-f(x)}{h}$

Put f(x) = 1/x3

So the derivative of 1/x3 from first principle is

$\dfrac{d}{dx}\big(\dfrac{1}{x^3} \big)$ $=\lim\limits_{h \to 0}\dfrac{\frac{1}{(x+h)^3}-\frac{1}{x^3}}{h}$

= limh→0 $\dfrac{x^3-(x+h)^3}{hx^3(x+h)^3}$

= limh→0 $\dfrac{x^3-(x^3+3x^2h+3xh^2+h^3)}{hx^3(x+h)^3}$

= limh→0 $\dfrac{-h(3x^2+3xh+h^2)}{hx^3(x+h)^3}$

= limh→0 $\dfrac{-(3x^2+3xh+h^2)}{x^3(x+h)^3}$

= $\dfrac{-(3x^2+0+0)}{x^3(x+0)^3}$

= $\dfrac{-3x^2}{x^6}$ = $-\dfrac{3}{x^4}$.

Thus, the derivative of 1/x3 is equal to -3/x4 and this is obtained from the first principle of derivatives.

Next, using the substitution method together with the product rule of derivatives, we will find the derivative of 1/x3. For this let us put

z=1/x3. We need to evaluate dz/dx.

This implies that

zx3=1

Differentiating with respect to x, we get that

$\dfrac{d}{dx}(zx^3)=\dfrac{d}{dx}(1)$

⇒ $z\dfrac{d}{dx}(x^3)+x^3\dfrac{d}{dx}(z)=0$ (by the product rule of derivatives)

⇒ z ⋅ 3x2 + x3 $\dfrac{dz}{dx}$ =0

⇒ x3 $\dfrac{dz}{dx}$ = -3zx2

⇒ $\dfrac{dz}{dx}=-\dfrac{3z}{x}$

⇒ $\dfrac{dz}{dx}=-\dfrac{3}{x^4}$ as z=1/x3

So we have obtained the differentiation of 1/x3 by product rule which is -3/x4.

Q1: Find the derivative of 1/x3.

Answer: The derivative of 1/x3 is -3/x4.

Q2: What is the derivative of x3?

Answer: The derivative of x3 is 3x2.

Share via: