D - dx 3 cos-6- x -4 cos 3-6- x -A- cos3 x - -60sin3 x - -60cos3 x -D- -60sin3 x -

The correct option is C π60cos (3x^∘)We know that cos 3x =4cos^3 x 3cos x Hence, ddx(3cos(π6+x^∘) 4cos^3 (π6+x^∘)) = d dx( cos 3(π6+x^∘)) = d dx( cos(π2+3x^∘ ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios of Allied Angles

d / dx 3 cosπ...

Question

$\frac{d}{d x} (3 cos (\frac{π}{6} + x^{\circ}) - 4 {cos}^{3} (\frac{π}{6} + x^{\circ})) =$

cos (3 x^{\circ})

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\frac{π}{60} sin (3 x^{\circ})

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\frac{π}{60} cos (3 x^{\circ})

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- \frac{π}{60} sin (3 x^{\circ})

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Solution

The correct option is C $\frac{π}{60} cos (3 x^{\circ})$
We know that $cos 3 x = 4 {cos}^{3} x - 3 cos x$
Hence, $\frac{d}{d x} (3 cos (\frac{π}{6} + x^{\circ}) - 4 {cos}^{3} (\frac{π}{6} + x^{\circ}))$

$= \frac{d}{d x} (- cos 3 (\frac{π}{6} + x^{\circ}))$

$= \frac{d}{d x} (- cos (\frac{π}{2} + 3 x^{\circ}))$

$= \frac{d}{d x} (sin 3 x^{\circ})$

$= \frac{d}{d x} (sin (3 x \times \frac{π}{180^{\circ}})) = \frac{3 π}{180^{\circ}} cos (3 x^{\circ})$

$= \frac{π}{60^{\circ}} cos (3 x^{\circ})$

Suggest Corrections

ddx(3cos(π6+x∘)−4cos3(π6+x∘))=

The correct option is C π60cos(3x∘)We know that cos3x=4cos3x−3cosx Hence, ddx(3cos(π6+x∘)−4cos3(π6+x∘)) =ddx(−cos3(π6+x∘)) =ddx(−cos(π2+3x∘)) =ddx(sin3x∘) =ddx(sin(3x×π180∘))=3π180∘cos(3x∘) =π60∘cos(3x∘)

$\frac{d}{d x} (3 cos (\frac{π}{6} + x^{\circ}) - 4 {cos}^{3} (\frac{π}{6} + x^{\circ})) =$