Question

$\left(\sin ^2\left(\theta \right)\right)'$

solve

Answer 1

$\left(\sin ^2\left(θ\right)\right)'\:=\sin \left(2θ\right)$

$\left(\sin ^2\left(θ\right)\right)'\:$

$\mathrm{Apply\:the\:chain\:rule}:\quad 2\sin \left(θ\right)\left(\sin \left(θ\right)\right)'\:$

$=2\sin \left(θ\right)\left(\sin \left(θ\right)\right)'\:$

$\left(\sin \left(θ\right)\right)'\:=\cos \left(θ\right)$

$=2\sin \left(θ\right)\cos \left(θ\right)$

$\mathrm{Use\:the\:following\:identity}:\quad \:2\cos \left(x\right)\sin \left(x\right)=\sin \left(2x\right)$

$=\sin \left(2θ\right)$

Plz consider taking 'I' as theta