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