Question

compute the derivative of sin x from the first principle
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Answer 1

Answer:

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Answer 2

Answer:

Step-by-step explanation:

We have,

$\tt{y=sin(x)}$

Now,

$\displaystyle\rm{\dfrac{dy}{dx}=\lim_{h\to0}\dfrac{sin(x+h)-sin(x)}{h}}$

Applying sin(C) - sin(D) formula,

$\displaystyle\rm{\implies\,\dfrac{dy}{dx}=\lim_{h\to0}\dfrac{2\,cos\left(\dfrac{2x+h}{2}\right)\,sin\left(\dfrac{x+h-x}{2}\right)}{h}}$

$\displaystyle\rm{\implies\,\dfrac{dy}{dx}=\lim_{h\to0}\dfrac{2\,cos\left(\dfrac{2x+h}{2}\right)\,sin\left(\dfrac{h}{2}\right)}{h}}$

$\displaystyle\rm{\implies\,\dfrac{dy}{dx}=\lim_{h\to0}\dfrac{cos\left(\dfrac{2x+h}{2}\right)\,sin\left(\dfrac{h}{2}\right)}{\dfrac{h}{2}}}$

$\displaystyle\rm{\implies\,\dfrac{dy}{dx}=\lim_{h\to0}\,cos\left(\dfrac{2x+h}{2}\right)\cdot\lim_{\frac{h}{2}\to0}\dfrac{\,sin\left(\dfrac{h}{2}\right)}{\dfrac{h}{2}}}$

$\displaystyle\rm{\implies\,\dfrac{dy}{dx}=cos\left(\dfrac{2x+0}{2}\right)\cdot1}$

$\displaystyle\rm{\implies\,\dfrac{dy}{dx}=cos\left(\dfrac{2x}{2}\right)}$

$\displaystyle\rm{\implies\,\dfrac{dy}{dx}=cos\left(x\right)}$