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Differentation formulas

Power rule​

f′(xn)=nxn−1f'(x^{n})=nx^{n-1}
Proof

The derivative of any function measures its instantaneous rate of change, defined as:

f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x)=\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}

For the power function, we get:

f′(x)=lim⁡h→0(x+h)n−xnhf'(x)=\lim_{h \to 0}\frac{(x+h)^{n}-x^{n}}{h}

If we simplify the numerator by expanding the brackets, we get:

(x+h)n=xn+nxn−1h+n(n−1)2xn−2h2+...+hn(x+h)^{n}=x^{n}+nx^{n-1}h+\frac{n(n-1)}{2}x^{n-2}h^{2}+...+h^{n}

If we substitute this simplified numerator back into the original formula:

f′(x)=lim⁡h→0xn+nxn−1h+n(n−1)2xn−2h2+...+hn−xnhf'(x)=\lim_{h \to 0}\frac{x^{n}+nx^{n-1}h+\frac{n(n-1)}{2}x^{n-2}h^{2}+...+h^{n}-x^{n}}{h}f′(x)=lim⁡h→0nxn−1h+n(n−1)2xn−2h2+...+hnhf'(x)=\lim_{h \to 0}\frac{nx^{n-1}h+\frac{n(n-1)}{2}x^{n-2}h^{2}+...+h^{n}}{h}

We then factor out every remaining term in the numerator and divide by h:

f′(x)=lim⁡h→0nxn−1+n(n−1)2xn−2h+...+hn−1f'(x)=\lim_{h \to 0} nx^{n-1}+\frac{n(n-1)}{2}x^{n-2}h+...+h^{n-1}

If we evaluate h to 0:

f′(x)=nxn−1+0+0+0...+0=nxn−1f'(x)=nx^{n-1} + 0 + 0 + 0 ... + 0 = nx^{n-1}