Question

Differentiation sin x with respect to x using first principle method.

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Answer 1

Topic :-

Differentiation - First principle method

To find :-

Differentiation of sin x

Solution :-

Let sin (x) = f(x)

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Apply first principle rule

• $\sf{f'(x)=\lim\limits_{h\to0} \dfrac{f(x+h)-f(x)}{h}}$

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Substitute f(x) = sin (x)

$\sf{\implies f'(x)=\lim\limits_{h\to0} \dfrac{\sin(x+h)-\sin(x)}{h}}$

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Apply trigonometric identity :-

• $\sf sin A - sin B = 2cos\dfrac{(A + B)}{2} . sin\dfrac{ (A - B)}2$

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$\sf{\implies f'(x)=\lim\limits_{h\to0} \dfrac{2\cos \left[\dfrac{(x+h)+x}{2}\right]\times \sin\left[\dfrac{(x+h)-x}{2}\right]}{h}}$

$\sf{\implies f'(x)=\lim\limits_{h\to0} \dfrac{2\cos \left[\dfrac{2x+h}{2}\right]\times \sin\left[\dfrac{h}{2}\right]}{h}}$

$\sf{\implies f'(x)=\lim\limits_{h\to0} \dfrac{\cos \left[\dfrac{2x+h}{2}\right]\times \sin\left[\dfrac{h}{2}\right]}{ \dfrac h 2}}$

$\sf{\implies f'(x)=\lim\limits_{h\to0} \left \{ \cos \left[\dfrac{2x+h}{2} \right] \right \}\times \lim \limits_{ h \to0} \left \{ \dfrac{\sin\left(\dfrac{h}{2}\right)}{ \left(\dfrac h 2 \right)} \right \}}$

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Apply formula:-

• $\sf \lim \limits_{x \to0} \dfrac{ \sin \: x}{x} = 1$

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$\sf{\implies f'(x)=\lim\limits_{h\to0} \left \{ \cos \left[\dfrac{2x+h}{2} \right] \right \}\times 1}$

$\sf{\implies f'(x)=\lim\limits_{h\to0} \cos \left[\dfrac{h}{2} + x \right] }$

Now directly substitute h = 0

$\sf{\implies f'(x)= \cos \left[\dfrac{0}{2} + x \right] }$

$\sf{\implies f'(x)= \cos \left( x \right) }$

Hence differentiation of sin(x) is cos(x).

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More :-

• $\sf \dfrac{d}{dx}(x)=1$

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• $\sf \dfrac{d}{dx}(x^n)=nx^{n-1}$

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• $\sf \dfrac{d}{dx}(\cos\:x)=-\sin\:x$

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• $\sf \dfrac{d}{dx}(\tan x)=\sec^2x$

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• $\sf \dfrac{d}{dx}(e^x)=e^x$

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• $\sf \dfrac{d}{dx}(\sec x)=\sec x . \tan x$