Question

find the derivative of sin(5x-8) with respect to first principle

Answer 1

for function f(x) , first derivative principle is given by
$\bold{Lim_{h \rightarrow0}\,\frac{f(x+h)-f(x)}{h}}$

here given f(x) = sin(5x - 8)
so, f(x + h) = sin(5x + 5h -8)

now, d(sin(5x -8))/dx = Lim_{h→0} {sin(5x + 5h -8) - sin(5x -8)}x

use formula,
sinC - sinD = 2cos(C + D)/2.sin(C - D)/2

then, Lim_{h→0} 2cos(5x + 5h/2 - 8).sin(5h/2)/h
now, take limit is in the form of 0/0 ,
we also know,
Lim_{x→0} sinax/x = a , use it here,

then, 2cos(5x - 8) Lim_{h→0} sin(5h/2)/h
= 2cos(5x - 8) × 5/2
= 5cos(5x - 8)

hence, d{sin(5x -8)}/dx = 5cos(5x -8)

Answer 2

Answer:

Step-by-step explanation: this question is solved in the attached sheet