Question
(Requires calculus) Use mathematical induction to prove that the derivative of $f(x)=x^n$ equals $n x^{n-1}$ whenever $n$ is a positive integer. (For the inductive step, use the product rule for derivatives.)
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Alternative proof of the Power Rule The Binomial Theorem states that for any positive integer $n$ $$\begin{aligned} (a+b)^{n}=& a^{n}+n a^{n-1} b+\frac{n(n-1)}{2 \cdot 1} a^{n-2} b^{2} \\ &+\frac{n(n-1)(n-2)}{3 \cdot 2 \cdot 1} a^{n-3} b^{3}+\cdots+n a b^{n-1}+b^{n} \end{aligned}$$ Use this formula and the definition $f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$ to show that $\frac{d}{d x}\left(x^{n}\right)=n x^{n-1},$ for any positive integer $n$
Derivatives
Rules of Differentiation
$$\begin{aligned} &\text { If } f(x)=x, \text { then } f^{\prime}(x)=1 \cdot x^{0}=1\\ &\text { If } f(x)=x^{2}, \text { then } f^{\prime}(x)=2 x^{1}=2 x \end{aligned}$$ (a) Show that if $\quad f(x)=x^{3}$ $$ \text { then } \quad f^{\prime}(x)=3 x^{2} $$ (b) Prove by induction that for each positive integer $n$ $$ f(x)=x^{n} \quad \text { has derivative } \quad f^{\prime}(x)=n x^{n-1} $$ HINT: $$ (x+h)^{k+1}-x^{k+1}=x(x+h)^{k}-x \cdot x^{k}+h(x+h)^{k} $$
The Derivative; The Process of Differentiation
The Derivative
If $f(x)=x,$ then $f^{\prime}(x)=1 \cdot x^{0}=1$ If $f(x)=x^{2},$ then $f^{\prime}(x)=2 x^{\prime}-2 x$ (a) Show that $$f(x)=x^{3}, \quad \text { then } \quad f^{\prime}(x)=3 x^{2}$$ (b) Prove by induction that for each positive integer $n$ $$f(x)=x^{n} \quad \text { has derivative } \quad f^{\prime}(x)=n x^{n-1}$$
The Derivative; The Process of Differentation
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