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%% Derivative %%

Power Rule ​

Tags
Cegep1
Mathematics
Word count
197 words
Reading time
2 minutes

Let f(x)=xn, n∈R, then

ddxxn=nxn−1

Proof ​

For now, this proof is only for $n \in \mathbb{N}$.

Note the fact that the following equality is true:

xn−an=(x−a)⋅∑i=1nxn−iai−1

Let f(x)=xn, we have to show that f′(a)=nan−1 for any a:

f′(a)=limh→0f(a+h)−f(a)h=limx→af(x)−f(a)x−a=limx→axn−anx−a

Using the fact stated above, we have

=limx→a(x−a)⋅∑i=1nxn−iai−1x−a=limx→a∑i=1nxn−iai−1=∑i=1nan−iai−1=∑i=1nan−1=nan−1

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