Use the first principle to differentiate? y=sin⁡x

Question

anita1415snck · Accepted Answer

Explanation:After you have your expression in that form, you can differentiate it using the Chain Rule:In your case: u12→12sin⁡(x)−12⋅ddxsin⁡(x)Then, 12sin⁡(x)−12⋅cos⁡(x) which is your answer

Videoad3u · Accepted Answer

Using the limit definition of the derivative we have:

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

So for the given function, where

f (x) = \sqrt{\sin x}

, we have:

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

= lim_{h \to 0} \frac{\sqrt{\sin (x + h)} - \sqrt{\sin x}}{h} \cdot \frac{\sqrt{\sin (x + h)} + \sqrt{\sin x}}{\sqrt{\sin (x + h)} + \sqrt{\sin x}}

= lim_{h \to 0} \frac{\sin (x + h) - \sin x}{h (\sqrt{\sin (x + h)} + \sqrt{\sin x})}

Then we can use the trigonometric identity:

\sin (A + B) \equiv \sin A \cos B + \cos A \sin B

Giving us:

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

= lim_{h \to 0} \frac{\sin x (\cos h - 1) + \cos x \sin h}{h (\sqrt{\sin (x + h)} + \sqrt{\sin x})}

= lim_{h \to 0} \frac{\cos h - 1}{h} \frac{\sin x}{\sqrt{\sin (x + h)} + \sqrt{\sin x}}

+ \frac{\sin h}{h} \frac{\cos x}{\sqrt{\sin (x + h)} + \sqrt{\sin x}}

Then we use two very standard calculus limits:

lim_{θ \to 0} \frac{\sin θ}{θ} = 1

, and

lim_{θ \to 0} \frac{\cos θ - 1}{θ} = 0

, and
And we can now evaluate the limits:

f^{'} (x) = 0 \times \frac{\sin x}{\sqrt{\sin (x)} + \sqrt{\sin x}} + 1 \times \frac{\cos x}{\sqrt{\sin (x)} + \sqrt{\sin x}}

= \frac{\cos x}{2 \sqrt{\sin (x)}}

Use the first principle to differentiate? \(\displaystyle{y}=\sqrt{{{\sin{{x}}}}}\)

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