Question

d/dx(9sin x +sin 3x) at x=π/3 is


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Answer 1

Given,

$\[\frac{{d\left( {9\sin x + \sin 3x} \right)}}{{dx}}\]$ at $\[x = \frac{\pi }{3}\]$

Solution,

$\[ = \frac{{d\left( {9\sin x + \sin 3x} \right)}}{{dx}}\]$

$\[ = 9\cos x + 3\cos 3x\]$

Put the given value of x.

$\[ = 9\cos \left( {\frac{\pi }{3}} \right) + 3\cos \left( {\frac{{3\pi }}{3}} \right)\]$

$\[ = 9\cos \left( {\frac{\pi }{3}} \right) + 3\cos \pi \]$

Simplify it,

$\[ = 9 \times \frac{1}{2} + 3 \times \left( { - 1} \right)\]$

$\[ = \frac{9}{2} - 3\]$

$\[ = \frac{3}{2}\]$

Hence the value is $\[\frac{3}{2}\]$.