Question

Find the derivative of cos x from first principle.

Answer 1

Let f (x) = cos x. Accordingly, from the first principle,
f'(x)=$\lim_{h\rightarrow 0}$ $\frac{f(x+h)-f(x)}{h}$
=$\lim_{h\rightarrow 0}$ $\frac{cos(x+h)-cos(x)}{h}$
=$\lim_{h\rightarrow 0}$ [$\frac{cos\,xcos\,h-sin\,xsin\,h-cos\,x}{h}$]
=$\lim_{h\rightarrow 0}$ [$\frac{-cos\,x(1-cos\,h)}{h}-\frac{sin\,xsin\,h}{h}$]
=-cos($\lim_{h\rightarrow 0}$ $\frac{1-cos\,h}{h}$) - sinx $\lim_{h\rightarrow 0}$($\frac{sin\,h}{h}$)
=-cosx(0 ) - sinx(1) [$\lim_{h\rightarrow 0}$ $\frac{1-cos\,h}{h}$ =0 and $\lim_{h\rightarrow 0}$ $\frac{sin\,h}{h}$ = 1]
=-sinx
∴f'(x) = -sinx