Question

Use the $z\to x$definition of the derivative to show that role="math" localid="1648386605099" $\frac{d}{dx}\left(x^8\right)=8x^7$.

Short answer

$$\begin{gathered} \underset{z\to x}{\lim }\frac{z^8-x^8}{z-x} \\ =\underset{z\to x}{\lim }\frac{(z-x)(z^7+x{z}^6+x^2{z}^5+x^3{z}^4+x^4{z}^3+x^5{z}^2+x^{6}z+x^7)}{z-x} \\ =\underset{z\to x}{\lim }(z^7+x{z}^6+x^2{z}^5+x^3{z}^4+x^4{z}^3+x^5{z}^2+x^{6}z+x^7) \\ =x^7+x^7+x^7+x^7+x^7+x^7+x^7+x^7 \\ =8x^7. \end{gathered}$$

Step-by-step solution

Step 1. Given Information 

We have given the following function :-

$x^8$.

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

$\frac{d}{dx}\left(x^8\right)=8x^7$

Step 2. Derivative of given function :- 

Consider the given function as :-

$f\left(x\right)=x^8$.

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 :-

role="math" localid="1648386886563" $\frac{d}{dx}\left(x^8\right)=\underset{z\to x}{\lim }\frac{z^8-x^8}{z-x}$

Simplify it :-

role="math" localid="1648386910268" $$\begin{gathered} \frac{d}{dx}\left(x^8\right)\underset{z\to x}{\lim }\frac{z^8-x^8}{z-x} \\ \Rightarrow \frac{d}{dx}\left(x^8\right)=\underset{z\to x}{\lim }\frac{(z-x)(z^7+x{z}^6+x^2{z}^5+x^3{z}^4+x^4{z}^3+x^5{z}^2+x^{6}z+x^7)}{z-x} \\ \Rightarrow \frac{d}{dx}\left(x^8\right)=\underset{z\to x}{\lim }(z^7+x{z}^6+x^2{z}^5+x^3{z}^4+x^4{z}^3+x^5{z}^2+x^{6}z+x^7) \\ \Rightarrow \frac{d}{dx}\left(x^8\right)=x^7+x^7+x^7+x^7+x^7+x^7+x^7+x^7 \\ \Rightarrow \frac{d}{dx}\left(x^8\right)=8x^7 \end{gathered}$$

Hence proved.