Chaya Galloway

2021-01-15

Find the derivative of the Function: $P\left(x\right)=\left(x-3\mathrm{cos}x\right)\left(x+3\mathrm{cos}x\right)$

dessinemoie

$P\left(x\right)=\left(x-3\mathrm{cos}x\right)\left(x+3\mathrm{cos}x\right)={x}^{2}-9{\mathrm{cos}}^{2}\left(x\right)$
$\frac{dp}{dx}=2x-9d\left({\mathrm{cos}}^{2}x\right)=2x-9\cdot 2\mathrm{cos}x\cdot d\left(\mathrm{cos}x\right)=$
$=2x-9\cdot 2\mathrm{cos}x\left(-\mathrm{sin}x\right)$
$=2x+9\cdot 2\mathrm{sin}x\mathrm{cos}x$
$=2x+9\mathrm{sin}\left(2x\right)$
you can use the formula
$\left(a-b\right)\left(a+b\right)={a}^{2}-{b}^{2}$
$2\mathrm{sin}x\mathrm{cos}x=\mathrm{sin}\left(2x\right)$
or you can use the product rule
$\frac{dp}{dx}=\left(d\left(x-3\mathrm{cos}x\right)\right)\left(x+3\mathrm{cos}x\right)+\left(x-3\mathrm{cos}x\right)\left(d\left(x+3\mathrm{cos}x\right)\right)=$
$\left(1+3\mathrm{sin}x\right)\left(x+3\mathrm{cos}x\right)+\left(x-3\mathrm{cos}x\right)\left(1-3\mathrm{sin}x\right)=$
$x+3\mathrm{cos}x+3x\mathrm{sin}x+9\mathrm{sin}x\mathrm{cos}x+x-3\mathrm{cos}x-3x\mathrm{sin}x+9\mathrm{sin}x\mathrm{cos}x=$
$2x+18\mathrm{sin}x\mathrm{cos}x=2x+9\cdot 2\mathrm{sin}x\mathrm{cos}x=2x+9\mathrm{sin}\left(2x\right)$

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