Double Math

# Proof of Power rule

Prove that:

\frac d{dx}\left(x\right)^n=nx^{n-1}\;\;\;\;\;\;\;\;\;\; ;n\in \mathbb{R}

y=x^n\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(1)

y+\delta y=\left(x+\delta x\right)^n\;\;\;\;\;\;\;\;\;\;(2)

Subtracting eq(1) from eq(2)

y+\delta y-y=\left(x+\delta x\right)^n-x^n

\delta y=\left(x+\delta x\right)^n-x^n

\delta y=x^n\left(\delta x\right)^0+\begin{pmatrix}n\\1\end{pmatrix}x^{n-1}\left(\delta x\right)^1+\begin{pmatrix}n\\2\end{pmatrix}x^{n-2}\left(\delta x\right)^2

+\begin{pmatrix}n\\3\end{pmatrix}x^{n-3}\left(\delta x\right)^2+\begin{pmatrix}n\\4\end{pmatrix}x^{n-4}+…….+\begin{pmatrix}n\\n\end{pmatrix}x^{n-n}\left(\delta x\right)^n-x^n

\delta y=\begin{pmatrix}n\\1\end{pmatrix}x^{n-1}\left(\delta x\right)^1+\begin{pmatrix}n\\2\end{pmatrix}x^{n-2}\left(\delta x\right)^2+\begin{pmatrix}n\\3\end{pmatrix}x^{n-3}\left(\delta x\right)^2+\begin{pmatrix}n\\4\end{pmatrix}x^{n-4}+…….+\begin{pmatrix}n\\n\end{pmatrix}x^{0}\left(\delta x\right)^n

\delta y=\delta x\left[\begin{pmatrix}n\\1\end{pmatrix}x^{n-1}+\begin{pmatrix}n\\2\end{pmatrix}x^{n-2}\left(\delta x\right)^1+\begin{pmatrix}n\\3\end{pmatrix}x^{n-3}\left(\delta x\right)^2+\begin{pmatrix}n\\4\end{pmatrix}x^{n-4}\left(\delta x\right)^3+…….+\begin{pmatrix}n\\n\end{pmatrix}\left(\delta x\right)^{n-1}\right].

\frac{\delta y}{\delta x}=\frac{\delta x\left[\begin{pmatrix}n\\1\end{pmatrix}x^{n-1}+\begin{pmatrix}n\\2\end{pmatrix}x^{n-2}\left(\delta x\right)^1+\begin{pmatrix}n\\3\end{pmatrix}x^{n-3}\left(\delta x\right)^2+\begin{pmatrix}n\\4\end{pmatrix}x^{n-4}\left(\delta x\right)^3+\dots\dots.+\begin{pmatrix}n\\n\end{pmatrix}\left(\delta x\right)^{n-1}\right]}{\delta x}

\frac{\delta y}{\delta x}=\left[\begin{pmatrix}n\\1\end{pmatrix}x^{n-1}+\begin{pmatrix}n\\2\end{pmatrix}x^{n-2}\left(\delta x\right)^1+\begin{pmatrix}n\\3\end{pmatrix}x^{n-3}\left(\delta x\right)^2+\begin{pmatrix}n\\4\end{pmatrix}x^{n-4}\left(\delta x\right)^3+\dots\dots.+\begin{pmatrix}n\\n\end{pmatrix}\left(\delta x\right)^{n-1}\right]

\underset{\delta x\rightarrow0}{lim}\frac{\delta y}{\delta x}=\underset{\delta x\rightarrow0}{lim}\left[\begin{pmatrix}n\\1\end{pmatrix}x^{n-1}+\begin{pmatrix}n\\2\end{pmatrix}x^{n-2}\left(\delta x\right)^1+\begin{pmatrix}n\\3\end{pmatrix}x^{n-3}\left(\delta x\right)^2+\begin{pmatrix}n\\4\end{pmatrix}x^{n-4}\left(\delta x\right)^3+\dots\dots.+\begin{pmatrix}n\\n\end{pmatrix}\left(\delta x\right)^{n-1}\right]

\frac{dy}{dx}=\left[\begin{pmatrix}n\\1\end{pmatrix}x^{n-1}+\begin{pmatrix}n\\2\end{pmatrix}x^{n-2}.0+\begin{pmatrix}n\\3\end{pmatrix}x^{n-3}.0+\begin{pmatrix}n\\4\end{pmatrix}x^{n-4}.0+\dots\dots.+\begin{pmatrix}n\\n\end{pmatrix}.0\right]

{\frac{dy}{dx}=\begin{pmatrix}n\\1\end{pmatrix}x^{n-1}}

\frac{dy}{dx}=\frac{n!}{\left(n-1\right){\displaystyle!}}x^{n-1}

\frac{dy}{dx}=\frac{n\left(n-1\right)!}{\left(n-1\right){\displaystyle!}}x^{n-1}

\boxed{\frac{dy}{dx}=nx^{n-1}}

The power rule for differentiation was developed independently by Isaac Newton and Gottfried Wilhelm Leibniz in the mid-17th century for rational power functions, which both were then used to derive the power rule for integrals as the inverse operation.

Here we have discussed the proof of power rule in the derivative. Power rule actually helps us to evaluate problems in an easy way.