find the derivetive of cos(2x+3) with respective x by 1 st principle method
Answers
Step-by-step explanation:
The derivative of cos(2x-3) from first principle is -2sin(2x-3)
From first principle, the derivative of a function is given by
f'(x)= \lim_{h \to \ 0} \frac{f(x+h)-f(x)}{h}f
′
(x)=lim
h→ 0
h
f(x+h)−f(x)
Here,
f(x) = cos(2x-3)
f(x+h) = cos(2x + 2h -3)
It's derivative from first principle is
\begin{gathered}f'(x) = \lim_{h \to \ 0} \frac{cos(2x+2h-3)-cos(2x-3)}{h} \\f'(x) = \lim_{h \to \ 0} \frac{-2sin(\frac{2x+2h-3+2x-3}{2})sin(\frac{2x+2h-3-2x+3}{2} ) }{h} \\cosC-cosD=-2sin(\frac{C+D}{2})sin( \frac{C-D}{2})\end{gathered}
f
′
(x)=
h→ 0
lim
h
cos(2x+2h−3)−cos(2x−3)
f
′
(x)=
h→ 0
lim
h
−2sin(
2
2x+2h−3+2x−3
)sin(
2
2x+2h−3−2x+3
)
cosC−cosD=−2sin(
2
C+D
)sin(
2
C−D
)
\begin{gathered}f'(x)=-2 \lim_{h \to \ 0} \frac{sin(2x+h-3)sinh}{h} \\f'(x)=-2 \lim_{h \to \ 0}sin(2x+h-3) \lim_{h \to \ 0}\frac{sinh}{h} \\f'(x)=-2sin(2x-3)\end{gathered}
f
′
(x)=−2
h→ 0
lim
h
sin(2x+h−3)sinh
f
′
(x)=−2
h→ 0
lim
sin(2x+h−3)
h→ 0
lim
h
sinh
f
′
(x)=−2sin(2x−3)
Hope this will help if it doesn't work try finding by couple of researches thank you
