Using limit definition of the derivative, show that (sin7x)' = 7cos7x. Thanks

Thursday, November 5, 2015

Using limit definition of the derivative, show that (sin7x)' = 7cos7x. Thanks

You need to use derivative definition to find what derivative of
function $\displaystyle{f{{\left({x}\right)}}}={\sin{{7}}}{x}$ is such that:

$\displaystyle{f{'}}{\left({x}\right)}=\lim_{{{h}-\gt{0}}}$
(f(x+h)-f(x))/h

$\displaystyle\lim_{{{h}-\gt{0}}}\frac{{{f{{\left({x}+{h}\right)}}}-{f{{\left({x}\right)}}}}}{{h}}=\lim_{{{h}-\gt{0}}}{\left({\sin{}}\right.}$
7(x+h)-sin 7x)/h

You need to substitute 0 for h such
that:

$\displaystyle\frac{{{\sin{{7}}}{x}-{\sin{{7}}}{x}}}{{0}}=\frac{{0}}{{0}}$

You
should use l'Hospital's theorem to solve the llimit such
that:

$\displaystyle\lim_{{{h}-\gt{0}}}\frac{{{\sin{{7}}}{\left({x}+{h}\right)}-{\sin{{7}}}{x}}}{{h}}=\lim_{{{h}-\gt{0}}}{\left({\left({\sin{}}\right.}\right.}$
7(x+h)-sin 7x)')/(h')

$\displaystyle\lim_{{{h}-\gt{0}}}\frac{{{\sin{{7}}}{\left({x}+{h}\right)}-{\sin{{7}}}{x}}}{{h}}=$
lim_(h-gt0) 7cos7(x+h)/1

Substituting 0 for h
yields:

$\displaystyle\lim_{{{h}-\gt{0}}}{7}{\cos{{7}}}\frac{{{x}+{h}}}{{1}}={7}{\cos{{7}}}\frac{{{x}+{0}}}{{1}}={7}{\cos{}}$
7x

Hence, evaluating the limit using definition of
derivative yields $\displaystyle{\left({\sin{{7}}}{x}\right)}'={7}{\cos{{7}}}{x}$ .

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Thursday, November 5, 2015