Question

derivative of sin4x by using first principle method ​

Answer 1

First Principle of Derivative:

If $f(x)$ be any function of $x,$ then its derivative is given by

$\displaystyle \frac{d}{dx}\{f(x)\}=f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$

Step-by-step explanation:

Let $f(x)=sin4x$

Then $f(x+h)-f(x)$

$=sin(4x+4h)-sin4x$

$=2\:cos(\frac{4x+4h+4x}{2})\:sin(\frac{4x+4h-4x}{2})$

$=2\:cos(4x+2h)\:sin2h$

Now $\displaystyle \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$

$\displaystyle =4\:\lim_{h\to 0}\frac{2\:cos(4x+2h)\:sin2h}{h}$

$\displaystyle =4\:\lim_{h\to 0}cos(4x+2h)\:\lim_{h\to 0}\frac{sin2h}{2h}$

$\displaystyle =4\:cos4x\:\times 1,\quad \because \lim_{h\to 0}\frac{sinh}{h}=1$

$=4\:cos4x$

Thus $\frac{d}{dx}(sin4x)=4\:cos4x$

Answer:

The required derivative is $4\:cos4x$.