Question

What derivative of sec2x by first principle

Answer 1

Step-by-step explanation:

you can understand by seeing the pic

Answer 2

The derivative of sec2x by the first principle is 2 sec2x tan2x.

Step-by-step explanation:

Given: sec2x

To Find: Derivative of sec2x by the first principle

Solution:

• Finding the derivative of sec2x by the first principle

By using the first principle, the derivative of sec2x is given as,

$\Rightarrow \dfrac{d}{dx} (sec(2x)) = \lim_{h \to 0} \Big[ \dfrac{sec(2(x+h)) -sec(2x) }{h} \Big]$

$\Rightarrow \dfrac{d}{dx} (sec(2x)) = \lim_{h \to 0} \dfrac{1}{h} \Big[ \dfrac{1}{cos(2(x+h))} - \dfrac{1}{cos(2x)} \Big]$

$\Rightarrow \dfrac{d}{dx} (sec(2x)) = \lim_{h \to 0} \dfrac{1}{h} \Big[ \dfrac{ cos(2x) - cos(2(x+h)}{cos(2(x+h))cos(2x)} \Big]$

Using the trigonometric identity $cosA - cosB = -2 sin (\frac{A+B}{2} )sin (\frac{A-B}{2} )$ in the above expression, we get,

$\Rightarrow \dfrac{d}{dx} (sec(2x)) = \lim_{h \to 0} \dfrac{1}{h} \Big[ \dfrac{ -2 sin(2x+h) sin(-h)}{cos(2(x+h))cos(2x)} \Big]$

$\Rightarrow \dfrac{d}{dx} (sec(2x)) = 2 \lim_{h \to 0} \Big( \dfrac{ sin(2x+h)}{cos(2(x+h))cos(2x)} \Big) \lim_{h \to 0} \Big( \dfrac{ sin(h)}{h} \Big)$

$\Rightarrow \dfrac{d}{dx} (sec(2x)) = 2 \Big( \dfrac{ sin(2x+0)}{cos(2(x+0))cos(2x)} \Big) \lim_{h \to 0} \Big( \dfrac{ sin(h)}{h} \Big)$

$\Rightarrow \dfrac{d}{dx} (sec(2x)) = 2 \Big( \dfrac{1}{cos(2x)} \times \dfrac{ sin(2x)}{cos(2x)} \Big) (1)$

$\Rightarrow \dfrac{d}{dx} (sec(2x)) = 2 sec(2x)} tan(2x)$

Hence, the derivative of sec2x by the first principle is 2 sec2x tan2x.