Question

Use the $z\to x$ definition of the derivative to show that $\frac{d}{dx}\left(x^4\right)=4x^3$.

Short answer

$$\begin{gathered} \underset{z\to x}{\lim }\frac{z^4-x^4}{z-x} \\ =\underset{z\to x}{\lim }\frac{(z-x)(z^3+x{z}^2+x^{2}z+x^3)}{z-x} \\ =\underset{z\to x}{\lim }(z^3+x{z}^2+x^{2}z+x^3) \\ =x^3+x\times x^2+x^2\times x+x^3 \\ =x^3+x^3+x^3+x^3 \\ =4x^3 \end{gathered}$$

Step-by-step solution

Step 1. Given Information

We have given the following function :-

$x^4$.

We have to use the $z\to x$definition of derivative, to prove that :-

$\frac{d}{dx}\left(x^4\right)=4x^3$

Step 2. Derivative of given function :-

Consider the given function as :-

$f\left(x\right)=x^4$.

We know that$z\to x$definition of derivative is stated that :-

$\frac{d}{dx}f\left(x\right)=\underset{z\to x}{\lim }\frac{f\left(z\right)-f\left(x\right)}{z-x}$

Put the values :-

localid="1648379528748" $\frac{d}{dx}\left(x^4\right)=\underset{z\to x}{\lim }\frac{z^4-x^4}{z-x}$

Simplify it :-

localid="1648379632597" $$\begin{gathered} \frac{d}{dx}\left(x^4\right)=\underset{z\to x}{\lim }\frac{z^4-x^4}{z-x} \\ \Rightarrow \frac{d}{dx}\left(x^4\right)=\underset{z\to x}{\lim }\frac{(z-x)(z^3+x{z}^2+x^{2}z+x^3)}{z-x} \\ \Rightarrow \frac{d}{dx}\left(x^4\right)=\underset{z\to x}{\lim }(z^3+x{z}^2+x^{2}z+x^3) \\ \Rightarrow \frac{d}{dx}\left(x^4\right)=x^3+x\times x^2+x^2\times x+x^3 \\ \Rightarrow \frac{d}{dx}\left(x^4\right)=x^3+x^3+x^3+x^3 \\ \Rightarrow \frac{d}{dx}\left(x^4\right)=4x^3 \end{gathered}$$

Hence proved.