Question

how do you prove that d/dx sinx is equal to cosx?

Answer 1

@NewyorkZ have you studied limits?

Answer 2

and differentiability?

Answer 3

yes

Answer 4

\[\frac{d}{dx}\sin x=\frac{\Delta}{\Delta x} \sin x\]
this can be written assuming sin x to be differentiable
\[\lim_{\Delta x \to 0} \frac{\sin (x+\Delta x)-\sin x}{x+\Delta x-x}\]

Answer 5

do you follow?

Answer 6

yes you are using the definition of limits eqaution f(x+h)-f(X)/h

Answer 7

can you expand this?
\[\sin (x+\Delta x)\]

Answer 8

sinx-1

Answer 9

Use this formula
\[\sin (A+B)=\sin A \cos B+\sin B\cos A\]

Answer 10

i am honestly lost

Answer 11

use the formula and expand
put
\[A=x\]
\[B=\Delta x\]

Answer 12

Following @ash2326 's lead
To expand
$$
\sin (x+\Delta x)
$$
we need
$$
\sin (A+B)=\sin A \cos B+\sin B\cos A
$$
But instead of \(A\) and \(B\), we need \(x\) and \(\Delta x\), respectively.

We need this because it's in the definition of derivative:
$$
\lim_{\Delta x \to 0} \frac{\sin (x+\Delta x)-\sin x}{x+\Delta x-x}
$$
So if you do this, you get
$$
\lim_{\Delta x \to 0} \frac{\sin x\cos\Delta x+\sin\Delta x+\sin\Delta x\cos x-\sin x}{x+\Delta x-x}
$$
Do you see now?

Answer 13

so the answer is 0 x sinx+cox (1)/1