how to solve derivative of sin3x using first principle - Maths - Limits and Derivatives

Question

how to solve derivative of sin3x using first
principle?

Vipin Verma · Accepted Answer

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

so f'(x) = $\displaystyle\lim_{{{x}\to{0}}}\frac{{{\sin{{3}}}{\left({x}+{h}\right)}-{\sin{{\left({3}{x}\right)}}}}}{{h}}$

$\displaystyle=\lim_{{{x}\to{0}}}\frac{{{2}{\sin{{\left(\frac{{{3}{x}+{3}{h}-{3}{x}}}{{2}}\right)}}}{\cos{{\left(\frac{{{3}{x}+{3}{h}+{3}{x}}}{{2}}\right)}}}}}{{h}}$

$\displaystyle=\lim_{{{x}\to{0}}}\frac{{{2}{\sin{{\left(\frac{{{3}{h}}}{{2}}\right)}}}{\cos{{\left(\frac{{{6}{x}+{h}}}{{2}}\right)}}}}}{{h}}$

$\displaystyle=\lim_{{{x}\to{0}}}{2}\frac{{{\sin{{\left(\frac{{{3}{h}}}{{2}}\right)}}}}}{{{\left(\frac{{{3}{h}}}{{2}}\right)}\cdot{\left(\frac{{2}}{{3}}\right)}}}{\cos{{\left(\frac{{{6}{x}+{h}}}{{2}}\right)}}}$

= 3cos3x
Its lim h -> 0, put h->0 in place of x -> 0

how to solve derivative of sin3x using first principle?

how to solve derivative of sin^3xusing first principle?