title | lang | category | permalink | ident | parent | kind | mathjax | layout | type |
---|---|---|---|---|---|---|---|---|---|
Derivatives of Polynomials |
en |
en |
en/theorem_derivatives_polynomials |
theorem_derivatives_polynomials |
derivatives |
theorem |
true |
post |
post |
Proof
$f(x)=c$;
$\displaystyle\lim_{h\longrightarrow 0}\dfrac{f(x+h)-f(x)}{h}=\lim_{h\longrightarrow 0}\dfrac{c-c}{h}=\lim_{h\longrightarrow 0} 0 = 0$
$\displaystyle\lim_{h\longrightarrow 0}\dfrac{f(x+h)-f(x)}{h}=\lim_{h\longrightarrow 0}\dfrac{c-c}{h}=\lim_{h\longrightarrow 0} 0 = 0$
Proof
$f(x)=x$;
$\displaystyle\lim_{h\longrightarrow 0}\dfrac{f(x+h)-f(x)}{h}=\lim_{h\longrightarrow 0}\dfrac{(x+h)-x}{h}=\lim_{h\longrightarrow 0} 1 = 1$
$\displaystyle\lim_{h\longrightarrow 0}\dfrac{f(x+h)-f(x)}{h}=\lim_{h\longrightarrow 0}\dfrac{(x+h)-x}{h}=\lim_{h\longrightarrow 0} 1 = 1$
Proof
$f(x)=x^2$;
$\displaystyle\lim_{h\longrightarrow 0}\dfrac{f(x+h)-f(x)}{h}=\lim_{h\longrightarrow 0}\dfrac{(x+h)^2-x^2}{h}=\\\displaystyle\lim_{h\longrightarrow 0} \dfrac{x^2+2xh+h^2-x^2}{h}=\lim_{h\longrightarrow 0} 2x+h = 2x$
$\displaystyle\lim_{h\longrightarrow 0}\dfrac{f(x+h)-f(x)}{h}=\lim_{h\longrightarrow 0}\dfrac{(x+h)^2-x^2}{h}=\\\displaystyle\lim_{h\longrightarrow 0} \dfrac{x^2+2xh+h^2-x^2}{h}=\lim_{h\longrightarrow 0} 2x+h = 2x$
Proof
$f(x)=x^n$;
According to the binomial expansion {% cite theorem_binomial_expansion %} <br><br>
$(x+h)^n=x^n+nx^{n-1}h+\binom{n}{2}x^{n-1}h^2+\cdots+\binom{n}{n-2}x^2h^{n-2}+nxh^{n-1}+h^n$ <br><br>
So <br><br>
$\displaystyle\lim_{h\longrightarrow 0}\dfrac{f(x+h)-f(x)}{h}=\lim_{h\longrightarrow 0}\dfrac{(x+h)^2-x^2}{h}=\\
\displaystyle\lim_{h\longrightarrow 0} \dfrac{x^n+nx^{n-1}h+\binom{n}{2}x^{n-1}h^2+\cdots+\binom{n}{n-2}x^2h^{n-2}+nxh^{n-1}+h^n-x^n}{h}=\\
\displaystyle\lim_{h\longrightarrow 0} nx^{n-1}+\binom{n}{2}x^{n-1}h+\cdots+\binom{n}{n-2}x^2h^{n-3}+nxh^{n-1}+h^{n-1} =\\
nx^{n-1}$