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iogann1982 [59]
2 years ago
7

0%5C%3A%202x%20%5C%3A%20%20Find%20%5Cfrac%7Bdy%7D%7Bdx%7D%20" id="TexFormula1" title=" \sf \large \: If \: y = Sin x \: * Cos \: 2x \: Find \frac{dy}{dx} " alt=" \sf \large \: If \: y = Sin x \: * Cos \: 2x \: Find \frac{dy}{dx} " align="absmiddle" class="latex-formula">
Thanku!​
Mathematics
2 answers:
DIA [1.3K]2 years ago
8 0

Answer:

  • cos (x) cos (2x) - 2sin(x) sin(2x)

Solution:

(See the solution in the photo)

dybincka [34]2 years ago
4 0

Answer:

\sf -5\cos \left(x\right)+6\cos ^3\left(x\right)

explanation:

\sf y  =  sin(x) * cos(2x)

\rightarrow \sf \frac{d}{dx}\left(sin\left(x\right)\ * \:\:cos\left(2x\right)\right)

\sf \bold {Apply\:the\:Product\:Rule}:\quad \left(f\cdot g\right)'=f\:'\cdot g+f\cdot g'

\rightarrow  \sf \frac{d}{dx}\left(\sin \left(x\right)\right)\cos \left(2x\right)+\frac{d}{dx}\left(\cos \left(2x\right)\right)\sin \left(x\right)

\sf \bold{ Apply \ differentiation \ rule \ \  \ : }   \ \ \ sin(x) = cos(x)  \ \ and  \ \ cos(x) = -sin(x)

\rightarrow  \sf \cos \left(x\right)\cos \left(2x\right)+\left(-\sin \left(2x\right)\ * \:2\right)\sin \left(x\right)

\rightarrow  \sf \cos \left(x\right)\cos \left(2x\right)\left-2\sin \left(2x\right)\sin \left(x\right)

\sf \bold {use \ the \ formulae \  : \  cos(2x) = 2cos^2(x) - 1} \ {and}  \ \ \sf \bold{sin(x) = 2 sin x cos x}

\rightarrow \sf cos(x) (2cos^2 (x) -1) -2(2sin(x)cos(x)sin(x))

\rightarrow \sf 2cos^3 (x) - cos(x) - 4sin^2(x) cos(x)

\rightarrow  \sf -5\cos \left(x\right)+6\cos ^3\left(x\right)

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