# Differentiation

In this page, we will come across proofs for some of the rules of differentiation which we use almost every time we come across a differentiation problem :

## The Power Rule

As per power rule for differentiation we know that

\[ \frac{δ(x^n) }{δx} = n\cdot x^{n-1} \]

(Above I have used greek word " δ " which means " a small change in " . )

Proof:

Let y = \( x^n \)

We will look at what kind of small change, we will get in y, if we change x by adding a small amount δx. Thus, we have :

\[ y = x^n \]

\[⇒ y+δy = {(x+δx)}^n \]

We can expand \( {(x+δx)} ^n\) using binomial expansion. This gives us

\( y+δy = x^n + n\cdot x^{n-1}(δx) + n(n-1)\cdot x^{n-2}({δx}^2)({2!}^{-1}) + higher powers of (δx) \)

Putting y = \( x^n \) and tidying up, we get

\( \frac{δy}{δx} = n\cdot x^{n-1}+ n(n-1)\cdot x^{n-2}(δx)({2!}^{-1}) + higher powers of (δx) \)

If we let δx → 0 , we get ,

\( \displaystyle \lim_{n \to 0} \frac{δy}{δx}= n\cdot x^{n-1} \)

(∵ everything except \(n\cdot x^{n-1}\) becomes so small that we can ignore it)

Hence, proved !!

## If y = \( ax^n \), find \( \frac{δy}{δx}\) .

We know from the above proof that \( \frac{δ(x^n) }{δx} = n\cdot x^{n-1} \). Therefore we have

\[ \frac{δ( ax^n )}{δx} = \frac {a\cdotδ( x^n )}{δx} \]

\[ ∴ \frac{ δ(x^n) }{δx} = an\cdot x^{n-1} \]

**Cite as:**Differentiation .

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/differentiation/