Prove that for any positive integer $n$, $\dfrac{d}{dx}(x^n) = nx^{n-1}$.

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Accepted Answer

>

> [!TLDR]
> Expanding $(x+h)^n$ by the binomial theorem and taking $h	o0$ gives $nx^{n-1}$.
>
> Let $f(x)=x^n$, $n$ a positive integer. By the first-principles (limit) definition of the derivative:
> $$f'(x) = \lim_{h	o 0} \frac{(x+h)^n - x^n}{h}$$
> Expand $(x+h)^n$ using the Binomial Theorem:
> $$(x+h)^n = x^n + nx^{n-1}h + \binom{n}{2}x^{n-2}h^2 + \cdots + h^n$$
> So
> $$(x+h)^n - x^n = nx^{n-1}h + \binom{n}{2}x^{n-2}h^2 + \cdots + h^n = h\left[nx^{n-1} + \binom{n}{2}x^{n-2}h + \cdots + h^{n-1}ight]$$
> Divide by $h$ ($h
e 0$): …

Q.Prove that for any positive integer $n$ , $\dfrac{d}{dx}(x^n) = nx^{n-1}$ .

Concept understanding — Rules of Differentiation

Q.Prove that for any positive integer nnn, ddx(xn)=nxn−1\dfrac{d}{dx}(x^n) = nx^{n-1}dxd​(xn)=nxn−1.

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Q.Prove that for any positive integer $n$ , $\dfrac{d}{dx}(x^n) = nx^{n-1}$ .

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