Question

find the derivative of the sec 3 X from the first principle with respect to X​

Answer 1

Step-by-step explanation:

we know cos3x=4cos*3x-3cosx

sec3x = 1/cos3x

d/dx(sec3x)=d/dx(1/cos3x)=d/dx(1/4cos*3x-3cosx)=

d/dx(1/4cos*3x)-d/dx(1/3cosx)

=1/4(3sec*2x.tanx)-1/3(secx.tanx)

Answer 2

$3sec3xtan3x$

Step-by-step explanation:

Let $f(x)=sec 3x$

f(x+h)=sec(3x+3h)

By first principle

$f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

Substitute the values

$f'(x)=\lim_{h\rightarrow 0}\frac{sec(3x+3h)-sec3x}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{\frac{1}{cos(3x+3h}-\frac{1}{cos3x}}{h}$

By using formula:$secx=\frac{1}{cosx}$

$f'(x)=\lim_{h\rightarrow 0}\frac{cos3x-cos(3x+3h)}{hcos(3x+3h)cos3x}$

$f'(x)=\lim_{h\rightarrow 0}\frac{-2sin(\frac{3x+3x+3h}{2})sin(\frac{3x-3x-3h}{2})}{hcos(3x+3h)cos3x}$

By using formula

$cosx-cosy=-2sin\frac{x+y}{2}sin\frac{x-y}{2}$

$f'(x)=\lim_{h\rightarrow 0}\frac{2sin(\frac{6x+3h}{2})sin(\frac{3h}{2})}{hcos(3x+3h)cos3x}$

By using sin(-x)=-sinx

$f'(x)=\lim_{h\rightarrow}\frac{2\times \frac{3}{2}sin\frac{6x+3h}{2}\frac{sin(\frac{3h}{2})}{\frac{3h}{2}}}{cos(3x+3h)cos3x}$

$f'(x)=\frac{3sin3x}{cos^2(3x)}$

By using $\lim_{h\rightarrow 0}\frac{sinh}{h}=1$

$f'(x)=3tan3xsec3x$

By using $tanx=\frac{sinx}{cosx},secx=\frac{1}{cosx}$

#Learns more:

https://brainly.in/question/12114469