Question

find derivative of sin2x at x= π/2

Answer 1

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Answer 2

Answer:

-2

Step-by-step explanation:

$f(x)=sin2x$

By using derivative formula:

$f'(a)= \lim_{h \to\ 0} \frac{f(a+h-f(a)}{h}$ (where h is small positive integer)

Now, derivative of $sin 2x$ at $x=\frac{\pi }{2}$

$f'(\frac{\pi }{2})= \lim_{h \to\ 0} \frac{f(\frac{\pi}{2} +h)-f( \frac{\pi }{2}) }{h}$

$= \lim_{h \to\ 0} \frac{sin 2(\frac{\pi}{2}+h )-sin2\frac{\pi}{2} }{h}$

$= \lim_{h \to\ 0} \frac{sin (\pi +2h )-sin\pi }{h}$

$= \lim_{h \to\ 0} \frac{-sin2h-0}{h}$

$=-\lim_{h \to\ 0} \frac{sin2h}{h}$

[Since it is of indeterminate form. We shall apply sandwich theorem to evaluate the limit.]

Multiply numerator and denominator by 2, we get

$=-\lim_{h \to\ 0} \frac{2sin2h}{2h}$

Using formula: $\lim_{h \to 0\frac{sinx}{x}=1$

$f'(\frac{\pi }{2})=-2 \times1=-2$

Derivative of $f(x)=sin2x$ at $\frac{\pi}{2}$ is $-2$.

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