Question

use the first principle to find the derivatives for f(x) = sec4x

Answer 1

f(x) = sec(4x)
f'(x) = 4 * sec(4x) * tan(4x)

Explanation,

take 4x = u

then 4 dx = du, or du/dx = 4

f(x) = y = sec(u)

dy/du = sec(u)*tan(u)

dy/dx = (dy/du) * (du/dx)

dy/dx = sec(u)*tan(u)*4

therefore,

dy/dx = 4*sec(4x)*tan(4x).

Answer 2

\[f \left( x \right)=\sec 4x ...(1)\]
\[f \left( x+\delta x \right)=\sec4 \left( x+\delta x \right) ...\left( 2 \right)\]
(2)-(1) gives
\[f \left( x+\delta x \right)-f \left( x \right)=\sec4 \left( x+\delta x \right)-\sec 4x\]
\[=\frac{ 1 }{ \cos \left( 4 x+4 \delta x \right) }-\frac{ 1 }{ \cos 4x }=\frac{ \cos 4 x-\cos \left( 4x+4 \delta x \right) }{ \cos \left( 4x+4 \delta x \right) \cos 4 x }\]
\[=\frac{ 2 \sin \frac{ 4x+4 x+4 \delta x }{ 2 }\sin \frac{ 4 x+4 \delta x-4 x }{ 2 } }{ \cos \left( 4x+4 \delta x \right) \cos 4 x }\]
\[=\frac{ 2\sin \left( 4x+2 \delta x \right)\sin 2 \delta x }{ \cos \left( 4x+4 \delta x \right) \cos 4 x }\]
divide by delta x
\[\frac{ f \left( x-\delta x \right)-f \left( x \right) }{ \delta x }=\frac{ 2 \sin \left( 4x+2 \delta x \right)\sin \frac{ 2 \delta x }{ \delta x } }{ \cos \left( 4 x+4 \delta x \right) \cos 4 x }\]
taking limits as delta x approaches 0
\[f \prime \left( x \right)=\lim_{x \rightarrow 0}\frac{ 2 \sin \left( 4x+2 \delta x \right)\frac{ \sin 2 \delta x }{ 2 \delta x }\times 2 }{ \cos \left( 4 x+4 \delta x \right)\cos 4 x }\]
\[=\frac{ 4 \sin 4x *1 }{ \cos 4 x \cos 4 x }=4 \sec 4x \tan 4 x\]

Answer 3

correction
write
\[\frac{ \sin 2 \delta x }{ \delta x } ~\in~place~of~\sin \frac{ 2 \delta x }{ \delta x }\]