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find dy/dx wheny= sin x sin 2x sin 3x sin 4x

Rinkoo Gupta
7 years ago
y=(sinxsin2x)(sin3xsin4x)
taking sinxsin2x as first funtion and sin3xsin4x as second function we apply th e rule of product of two function
dy/dx=sinxsin2x.{sin3x.4cos4x+sin4x3cos3x}+(sin3xsin4x){sinx.2cos2x+sin2xcosx}
=4sinxsin2xsin3xcos4x+3sinxsin2xsin4xcos3x+2sin3xsin4xsinxcos2x+sin3xsin4xsin2xcosx
Thanks & Regards
Rinkoo Gupta
Arun Kumar IIT Delhi
7 years ago
Hello
$\\ \frac{d}{dx}\left(\sin \left(x\right)\sin \left(2x\right)\sin \left(3x\right)\sin \left(4x\right)\right) \\ f=\sin \left(x\right),\:g=\sin \left(2x\right)\sin \left(3x\right)\sin \left(4x\right) \\ =\frac{d}{dx}\left(\sin \left(x\right)\right)\sin \left(2x\right)\sin \left(3x\right)\sin \left(4x\right)+\frac{d}{dx}\left(\sin \left(2x\right)\sin \left(3x\right)\sin \left(4x\right)\right)\sin \left(x\right) \\ \frac{d}{dx}\left(\sin \left(x\right)\right) \\ =\cos \left(x\right) \\ \frac{d}{dx}\left(\sin \left(2x\right)\sin \left(3x\right)\sin \left(4x\right)\right)$
$\\ =\cos \left(x\right)\sin \left(2x\right)\sin \left(3x\right)\sin \left(4x\right)+\left(\cos \left(2x\right)2\sin \left(3x\right)\sin \left(4x\right)+\left(\cos \left(3x\right)3\sin \left(4x\right)+\cos \left(4x\right)4\sin \left(3x\right)\right)\sin \left(2x\right)\right)\sin \left(x\right) \\=>2\sin \left(3x\right)\sin \left(4x\right)\cos \left(2x\right)\sin \left(x\right)+\sin \left(2x\right)\left(3\sin \left(4x\right)\cos \left(3x\right)\sin \left(x\right)+\sin \left(3x\right)\left(\sin \left(4x\right)\cos \left(x\right)+4\cos \left(4x\right)\sin \left(x\right)\right)\right)$
Thanks & Regards
Arun Kumar
Btech, IIT Delhi