+9673 We first work out the derivative of sinx cosx
Using product rule:
\(\dfrac{d}{dx}\sin x \cos x \\=(\cos x)(\cos x)+(-\sin x)(\sin x)\\=\cos 2x\)
Using chain rule:
\(y = \sqrt u, u=\sin x\cos x\)
\(\dfrac{dy}{dx}=\dfrac{dy}{du}\times \dfrac{du}{dx}=\dfrac{1}{2\sqrt{u}}\times \cos 2x=\dfrac{\cos 2x}{2\sqrt{\sin x\cos x}}\)
+9673 We first work out the derivative of sinx cosx
Using product rule:
\(\dfrac{d}{dx}\sin x \cos x \\=(\cos x)(\cos x)+(-\sin x)(\sin x)\\=\cos 2x\)
Using chain rule:
\(y = \sqrt u, u=\sin x\cos x\)
\(\dfrac{dy}{dx}=\dfrac{dy}{du}\times \dfrac{du}{dx}=\dfrac{1}{2\sqrt{u}}\times \cos 2x=\dfrac{\cos 2x}{2\sqrt{\sin x\cos x}}\)
+118687 Thanks Max, I am just going to do it a different way.
What's the derivative of \(\sqrt{\sin x \cos x}\;?\)
\(sin2x=2\sin x \cos x\\so\\ 0.5sin2x=\sin x \cos x\\ \frac{d}{dx}\;[0.5sin2x]^{0.5}\\ =\frac{d}{dx}\;\sqrt{0.5} \;\;[sin2x]^{0.5}\\ =\;\sqrt{0.5}*0.5 \;\;[sin2x]^{-0.5}*2cos(2x)\\ =\;\sqrt{0.5} \;\;\frac{cos(2x)}{ \sqrt{sin(2x)} }\\ = \;\;\frac{cos(2x)}{ \sqrt{2sin(2x)} }\\ = \;\;\frac{cos(2x)}{\sqrt{2*2sinxcosc} }\\ = \;\;\frac{cos(2x)}{2\sqrt{sinxcosc} }\\ \)
There you go - our answers are the same ... hopefully that means we are both correct !!!
