Question

How will u prove that derivative of sin x = cos x?​

Answer 1

To prove :-

Derivative of $\rm sin \,x = cos\,x$

Solution :-

Let's assume that $\rm f(x) = sin (x)$

Now according to the first principles,

$\rm\implies {\bf{f'(x) = \lim\limits_{h\to 0}\dfrac{f(x + h) - f(x) }{h}}}$

$\rm\implies f'(x) = \lim\limits_{h\to 0}\dfrac{sin(x + h) - sin(x) }{h}$

Apply formula,

$\bf sin \,A - sin\,B = 2\, cos\,\left(\dfrac{A+B}{2}\right).sin\left(\dfrac{A-B}{2}\right)$

$\rm\implies f'(x) = \lim\limits_{h\to 0}\dfrac{2\, cos\,\left(\dfrac{(x+h) + (x)}{2}\right).sin\left(\dfrac{(x+h)-(x)}{2}\right)}{h}$

$\rm\implies f'(x) = \lim\limits_{h\to 0}\dfrac{2\, cos\,\left(\dfrac{x+h+x}{2}\right).sin\left(\dfrac{x+h-x}{2}\right)}{h}$

$\rm\implies f'(x) = \lim\limits_{h\to 0}\dfrac{2\, cos\,\left(\dfrac{2x+h}{2}\right).sin\left(\dfrac{h}{2}\right)}{h}$

$\rm\implies f'(x) = \lim\limits_{h\to 0}\dfrac{\, cos\,\left(\dfrac{2x+h}{2}\right).sin\left(\dfrac{h}{2}\right)}{\left(\dfrac {h}{ 2} \right)}$

$\rm\implies f'(x) = \lim\limits_{h\to0}\left\{ cos\,\left(\dfrac{2x+h}{2}\right) \right\}\times \lim\limits_{h\to 0}\left\{\dfrac{sin\left(\dfrac{h}{2}\right)}{\left(\dfrac {h}{ 2} \right)}\right\}$

Here, since h is tending towards 0, h/2 will also tends to 0.

$\rm\implies f'(x) = \lim\limits_{h\to0}\left\{ cos\,\left(\dfrac{2x+h}{2}\right) \right\}\times \lim\limits_{h/2\to 0}\left\{\dfrac{sin\left(\dfrac{h}{2}\right)}{\left(\dfrac {h}{ 2} \right)}\right\}$

We are aware about the formula,

$\bf\lim\limits_{x\to0}\dfrac{sin\,x}{x} = 1$

Applying this, we get :

$\rm\implies f'(x) = \lim\limits_{h\to0}\left\{ cos\,\left(\dfrac{2x+h}{2}\right) \right\}\times {\bf{\lim\limits_{h\to 0}\left\{\dfrac{sin\left(\dfrac{h}{2}\right)}{\left(\dfrac {h}{ 2} \right)}\right\}}}$

$\rm\implies f'(x) = \lim\limits_{h\to0}\left\{ cos\,\left(\dfrac{2x+h}{2}\right) \right\}\times {\bf{1}}$

$\rm\implies f'(x) = \lim\limits_{h\to0}\left\{ cos\,\left(\dfrac{2x+h}{2}\right) \right\}$

$\rm\implies f'(x) = \lim\limits_{h\to0}\left\{ cos\,\left(\dfrac{2x}{2} + \dfrac{h}{2}\right) \right\}$

$\rm\implies f'(x) = \lim\limits_{h\to0}\left\{ cos\,\left(x + \dfrac{h}{2}\right) \right\}$

Now, solving the limit by direct substituition method :

$\rm\implies f'(x) = \lim\limits_{h\to0}\left\{ cos\,\left(x + \dfrac{h}{2}\right) \right\}$

$\rm\implies f'(x) =\left\{ cos\,\left(x + \dfrac{0}{2}\right) \right\}$

$\rm\implies f'(x) =\left\{ cos\,\left(x + 0\right) \right\}$

$\rm\implies f'(x) =cos\,\left(x\right)$

Hence derivative of $\rm sin\,x = cos\,x$