Find the derivative of the functions from first principle: $$cos(x-\dfrac{\pi}{8})$$

Asked by Pragya Singh | 1 year ago |  92

##### Solution :-

Let f(x) =$$cos(x-\dfrac{\pi}{8})$$ Accordingly, f(x + h) = cos 8

Accordingly, f(x + h)

$$cos(x+h-\dfrac{\pi}{8})$$

By first principle,

$$f'(x)=\lim\limits_{h \to 0}\dfrac{f(x+h)-f(x)}{h}$$

$$\lim\limits_{h \to 0} \dfrac{1}{h}[cos(x+h-\dfrac{\pi}{8})$$

$$-cos(\pi-\dfrac{\pi}{8})]$$ $$\lim\limits_{h \to 0} \dfrac{1}{h}[-2sin(\dfrac{2x+h-\dfrac{\pi}{4}}{2})sin(\dfrac{h}{2})]$$

$$[-sin(\dfrac{2x+h-\dfrac{\pi}{4})}{2}\dfrac{sin(\dfrac{h}{2})}{(\dfrac{h}{2})}]$$ $$-sin( \dfrac{2x+0-\dfrac{\pi}{4}}{2}).1$$

$$sin(x-\dfrac{\pi}{8})$$

Answered by Pragya Singh | 1 year ago

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