I am new to Calculus. rate of change

[Abu Rehan]

I am new to Calculus. I know that the rate of change in some thing is called it's derivative or we differentiate something to find the rate of change in it. But while differentiating or better to say proving that if f(x)= sinx then f'(x)= cosx, we take f'(x)= lim h->0 [sin(x+h)- sinx]/h. Can you explain why?
I know formulas like sin²x+ cos²x= 1
And the rest used in proving this.

 

[AZW]

Well to find the derivative of f(x) you find the the slope of the tangent line at every point x. To do this imagine a secant line intersecting the function at 2 points, (x,f(x)), and (x+h, f(x+h)). Finding the slope of the secant line, which is [f(x+h) - f(x)]/h, will approximately give you how the function is changing, depending on how small your "h" is. To find the tangent line, which only intersects the function at point (x, f(x)), simply let "h" go to zero. That gives you your formula for differentiation.

 

[Bohrok]

I am new to Calculus. I know that the rate of change in some thing is called it's derivative or we differentiate something to find the rate of change in it. But while differentiating or better to say proving that if f(x)= sinx then f'(x)= cosx, we take f'(x)= lim h->0 [sin(x+h)- sinx]/h. Can you explain why?
I know formulas like sin²x+ cos²x= 1
And the rest used in proving this.

Do you mean you want to know why, if f(x) = sinx,

$$f'(x) = \lim_{h\rightarrow 0} \frac{\sin(x + h) - \sin(x)}{h} = \cos x ?$$

You would first start by expanding sin(x + h) using the sum formula for sine, rewrite things a bit, then use some trig limits to evaluate the limit for the derivative.