differentiate-each-function-from-first-principle-1-f-x-1-1-x-2-f-x-1-2x-3-3-f-x-sin2x-4-f-x-co2x-

Question Number 11920 by carrot last updated on 05/Apr/17

d i f f e r e n t i a t e e a c h f u n c t i o n f r o m f i r s t p r i n c i p l e

1) f (x) = 1 + \frac{1}{x}

2) f (x) = \frac{1}{2 x + 3}

3) f (x) = s i n 2 x

4) f (x) = c o 2 x

Answered by sandy_suhendra last updated on 05/Apr/17

1) f (x) = 1 + x^{- 1}

f^{'} (x) = - x^{- 2} = \frac{- 1}{x^{2}}

2) f (x) = {(2 x + 3)}^{- 1}

f^{'} (x) = - 1 {(2 x + 3)}^{- 2} (2) = \frac{- 2}{{(2 x + 3)}^{2}}

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

4) f^{'} (x) = - 2 \sin 2 x

Commented by carrot last updated on 05/Apr/17

o h w h a t y o u d i d w a s d i f f e r e n t i a t e i t a n d t h a t s n o t w h a t t h e y a s k e d f o r

Commented by sandy_suhendra last updated on 06/Apr/17

oh sorry because I {don}^{'} t understand with the question

but I have fixed for no 3 and 4

Answered by FilupS last updated on 05/Apr/17

(1)

f^{'} (x) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h}

f^{'} (x) = \lim_{h \to 0} \frac{(1 + \frac{1}{x + h}) - (1 + \frac{1}{x})}{h}

= \lim_{h \to 0} \frac{\frac{1}{x + h} - \frac{1}{x}}{h}

= \lim_{h \to 0} \frac{(\frac{x - x - h}{x^{2} + x h})}{h}

= \lim_{h \to 0} \frac{(\frac{- h}{x^{2} + x h})}{h}

= \lim_{h \to 0} (- \frac{1}{x^{2} + x h})

= - \frac{1}{x^{2}}

Answered by FilupS last updated on 05/Apr/17

(2)

f^{'} (x) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h}

f^{'} (x) = \lim_{h \to 0} \frac{(\frac{1}{2 x + 2 h + 3}) - (\frac{1}{2 x + 3})}{h}

= \lim_{h \to 0} \frac{(\frac{1}{2 x + 2 h + 3} - \frac{1}{2 x + 3})}{h}

= \lim_{h \to 0} \frac{(\frac{2 x + 3 - 2 x - 2 h - 3}{(2 x + 2 h + 3) (2 x + 3)})}{h}

= \lim_{h \to 0} \frac{(\frac{- 2 h}{(4 x^{2} + 6 x + 4 h x + 6 h + 6 x + 9)})}{h}

= \lim_{h \to 0} \frac{(\frac{- 2}{(4 x^{2} + 6 x + 4 h x + 6 h + 6 x + 9)})}{1}

= - \frac{2}{4 x^{2} + 12 x + 9}

= - \frac{2}{{(2 x + 3)}^{2}}

Answered by sandy_suhendra last updated on 06/Apr/17

3) li \underset{h \to 0}{m} \frac{\sin 2 (x + h) - \sin 2 x}{h}

= li \underset{h \to 0}{m} \frac{2 \cos (2 x + h) \sin h}{h}

= li \underset{h \to 0}{m} 2 \cos (2 x + h) \frac{\sin h}{h}

= 2 \cos (2 x + 0) . 1 = 2 \cos 2 x

4) li \underset{h \to 0}{m} \frac{\cos 2 (x + h) - \cos 2 x}{h}

= li \underset{x \to 0}{m} \frac{- 2 \sin (2 x + h) \sin h}{h}

= li \underset{x \to 0}{m} - 2 \sin (2 x + h) \frac{\sin h}{h}

= - 2 \sin (2 x + 0) . 1 = - 2 \sin 2 x