Question

Find the derivative of sin x from first principle

Answer 1

i hope it will help u ..

Answer 2

Concept:-

A derivative is a contract between two or more parties.

Given:-

$sin x$

Find:-

Find the derivative of sin x from the first principle.

Solution:-

According to the problem, $f(x)=sinx$.

By definition of the derivative, $f'(x)= \lim_{h\to \00} \frac{f(x+h)-f(x)}{h}$

∵ $f'(x)= \lim_{h\to \00} \frac{sin(x+h)-sinx}{h}$

Using $sin(A+B)=sinAcosB+sinBcosA$ we get

$f'(x)= \lim_{h \to \00} \frac{sinxcosh+sinhcosx-sinx}{h}$

$f'(x)= \lim_{h \to \00} \frac{sinx(cosh-1)+sinhcosx}{h}$

$f'(x)= \lim_{h \to \00} ( \frac{sinx(cosh-1)}{h}+\frac{sinhcosx}{h})$

$f'(x)= \lim_{h \to \00} \frac{sinx(cosh-1)}{h}+ \lim_{h \to \00} \frac{sinhcosx}{h}$$f'(x)=sinx \lim_{h \to \00} \frac{(cosh-1)}{h}+ cosx \lim_{h \to \00} \frac{sinh}{h}$

We know have to rely on some standard limits:

$\lim_{h \to \00} \frac{sinh}{h}$= $1$

$\lim_{h\to \00} \frac{cosh - 1}{h}$=$0$

And so using these we have:

$f'(x)=0+(cosx)(1)=cosx$

hence,

$\frac{d}{dx}sinx=cosx$