Why is this true?
Differentiation of xn
f(x) = xn
f'(x) = nx(n-1)
Can someone explain? Thanks.
+9519 Use the first principles.
\(\dfrac{d}{dx}x^n\\ =\displaystyle\lim_{h\rightarrow 0}\dfrac{(x+h)^n-x^n}{h}\\ =\displaystyle\lim_{h\rightarrow 0}\dfrac{x^n+\binom{n}{1}x^{n-1}h+...-x^n}{h}\\\text{other terms are negligible as they evaluate to 0 after taking limit.}\\ =\displaystyle \lim_{h\rightarrow 0}\dfrac{nhx^{n-1}}{h} + \text{lots of negligible terms which evaluates to 0}\\ =\displaystyle \lim_{h\rightarrow 0}nx^{n-1}\\ =nx^{n-1}\text{ YAY}\)
+26367 Simple rules of differentiation
Why is this true?
Differentiation of xn
f(x) = xn
f'(x) = nx(n-1)
difference quotient:
\(\begin{array}{|rcll|} \hline \frac{\Delta y}{\Delta x} &=& \frac{(x+h)^n-x^n}{h} \\ \frac{\Delta y}{\Delta x} &=& \frac{ \binom n0 x^n + \binom n1 x^{n-1}h+ \binom n2 x^{n-2}h^2\ldots + \binom nn h^n -x^n }{h} \quad & | \quad \binom n0 = 1 \\ \frac{\Delta y}{\Delta x} &=& \frac{ x^n + \binom n1 x^{n-1}h+ \binom n2 x^{n-2}h^2\ldots + \binom nn h^n -x^n }{h} \\ \frac{\Delta y}{\Delta x} &=& \frac{ \binom n1 x^{n-1}h+ \binom n2 x^{n-2}h^2\ldots + \binom nn h^n }{h} \\ \frac{\Delta y}{\Delta x} &=& \binom n1 x^{n-1}+ \binom n2 x^{n-2}h^1\ldots + \binom nn h^{n-1} \quad & | \quad \binom n1 = n \\ \frac{\Delta y}{\Delta x} &=& n x^{n-1}+ \binom n2 x^{n-2}h\ldots + \binom nn h^{n-1} \\ \hline \end{array} \)
differential quotient:
\(\begin{array}{|rcll|} \hline \frac{\delta y}{\delta x} =f'(x) &=& \lim \limits_{h\to 0} \left( n x^{n-1}+ \binom n2 x^{n-2}h\ldots + \binom nn h^{n-1} \right) \\ f'(x) &=& n x^{n-1}+ \binom n2 x^{n-2}\times 0 \ldots + \binom nn \times 0^{n-1} \\ f'(x) &=& n x^{n-1} + 0 \ldots +0 \\ f'(x) &=& n x^{n-1} \\ \hline \end{array} \)
