Calculus derivatives

[Robin123]

Can you please send me the solution about how to find the derivative of sec^2x by first principle method without using l'hopital rule please. Send me the solution.

[Guest]

Why? How would it help you?

[Guest]

Since this has been here for a while:

sec(x)= 1/cos(x). sec(x+h)= 1/cos(x+ h).

sec(x+h)- sec(x)= 1/cos(x+h)- 1/cos(x)= (cos(x)- cos(x+h))/cos(x)cos(x+h).

So $\frac{sec(x+h)- sec(x)}{h}= \frac{cos(x)- cos(x+h)}{h cos(x)cos(x+h)}$.

Now use the trig identity cos(a+ b)= cos(a)cos(b)- sin(a)si(b):
cos(x+h)= cos(x)cos(h)- sin(x)sin(h)

$\frac{cos(x)- cos(x)cos(h)+ sin(x)sin(h)}{h cos(x)cos(x+h)}= \frac{cos(x)(1- cos(h))}{h cos(x)cos(x+h)}+ \frac{sin(x)sin(h)}{h cos(x)cos(x+h)}$

To finish that use $\lim_{h\to 0} \frac{1- cos(h)}{h}= 0$ and $\lim_{h\to 0}\frac{sin(h)}{h}= 1$.