Question

Find from first principles the derivative of cos x​

Answer 1

Answer:

Please see the explanation.

Step-by-step explanation:

Let

$f\left(x\right)=cosx$

By the first principle

$f\:'\left(x\right)=\lim _{h\to 0}\left(\frac{f\left(x+h\right)-f\left(x\right)}{h}\right)$

          $=\lim _{h\to 0}\left(\frac{cos\:\left(x+h\right)-cos\:x}{h}\right)$

           $=\lim _{h\to 0}\left[\frac{cos\:x\:cos\:h-sin\:x\:sin\:h\:-\:cos\:x}{h}\right]$

           $=\lim _{h\to 0}\left[\frac{-cos\:x\left(1-cos\:h\right)-sin\:x\:sin\:h\:}{h}\right]$

           $=\lim _{h\to 0}\left[\frac{-cos\:x\left(1-cos\:h\right)\:}{h}-\frac{sin\:x\:sin\:h}{h}\right]$

           $=-cosx\:\left(\lim \:_{h\to \:0\:}\frac{1-cos\:h}{h}\right)-sin\:x\:\lim \:\:_{h\to \:\:0}\:\left(\frac{sin\:h}{h}\right)$

           $=-cosx\:\left(0\right)-sinx\left(1\right)$

            $=-sin\:x$