Question

Find d/dx sin³x(4x-1) cos³(2x+3)

Answer 1

Answer:

$cos^{3}(2x+3)[3sin^{2}x(4x-1)cosx(4x-1)(8x-1)]-[6sin^{3}x(4x-1)cos^{2}(2x-1)sin(2x+3)]$

Step-by-step explanation:

using product rule:

=$\frac{\mathrm{d} }{\mathrm{d} x}[sin^{3}x(4x-1)]\times cos^{3}(2x+3)+sin^{3}x(4x-1)\frac{\mathrm{d} }{\mathrm{d} x}cos^3(2x+3)$

=$cos^{3}(2x+3)[3sin^{2}x(4x-1)cosx(4x-1)(8x-1)]+sin^{3}x(4x-1)3cos^{2}(2x+3)(-sin(2x+3))(2)$

Answer 2

Answer:

$3sin^{2}x\left ( 4x-1 \right )\cdot cosx\left ( 4x-1 \right )\cdot \left ( 8x-1 \right )\cdot cos^{3}\left ( 2x+3 \right )-6sin^{3}x\left ( 4x-1 \right )cos^{2}\left ( 2x+3 \right )\cdot sin\left ( 2x+3 \right )$

Step-by-step explanation:

$\frac{\mathrm{d} \left ( sin^{3}x\left ( 4x-1 \right )cos^{3} \left ( 2x+3 \right )\right )}{\mathrm{d} x}$

By using product rule :

$\frac{\mathrm{d} \left ( u\cdot v \right )}{\mathrm{d} x}$

$\frac{\mathrm{d} u}{\mathrm{d} x}\cdot v+u\cdot \frac{\mathrm{d} v}{\mathrm{d} x}$

therefore,

$\frac{\mathrm{d} \left ( sin^{3}x\left ( 4x-1 \right )cos^{3} \left ( 2x+3 \right )\right )}{\mathrm{d} x}$

$=[3sin^{2}x\left ( 4x-1 \right )\cdot cosx\left ( 4x-1 \right )\cdot \left ( 8x-1 \right )]cos^{3}\left ( 2x+3 \right )+sin^{3}x\left ( 4x-1 \right )[3cos^{2}\left ( 2x+3 \right )\cdot -sin\left ( 2x+3 \right )\cdot 2]$

$=3sin^{2}x\left ( 4x-1 \right )\cdot cosx\left ( 4x-1 \right )\cdot \left ( 8x-1 \right )\cdot cos^{3}\left ( 2x+3 \right )-6sin^{3}x\left ( 4x-1 \right )cos^{2}\left ( 2x+3 \right )\cdot sin\left ( 2x+3 \right )$