Solution 1.2:1a

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Because the expression is a product of two factors, we use the product rule:
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Because the expression is a product of two factors, we use the product rule,
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{{Displayed math||<math>\begin{align}
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(\sin x\cdot\cos x)^{\prime }
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&= (\cos x)^{\prime }\cdot\sin x + \cos x\cdot (\sin x)^{\prime }\\[5pt]
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&= -\sin x\cdot\sin x + \cos x\cdot\cos x\\[5pt]
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&= -\sin^2\!x + \cos^2\!x\,\textrm{.}
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\end{align}</math>}}
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<math>\begin{align}
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Using the formula for double angles, the answer can be simplified to <math>\cos 2x\,</math>.
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& \left( \sin x\centerdot \cos x \right)^{\prime }=\left( \cos x \right)^{\prime }\centerdot \sin x+\cos x\centerdot \left( \sin x \right)^{\prime } \\
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& \\
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& =-\sin x\centerdot \sin x+\cos x\centerdot \cos x=\sin ^{2}x+\cos ^{2}x \\
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\end{align}</math>
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Using the formula for double angles, the answer can be simplified to
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<math>\cos 2x</math>
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Current revision

Because the expression is a product of two factors, we use the product rule,

\displaystyle \begin{align}

(\sin x\cdot\cos x)^{\prime } &= (\cos x)^{\prime }\cdot\sin x + \cos x\cdot (\sin x)^{\prime }\\[5pt] &= -\sin x\cdot\sin x + \cos x\cdot\cos x\\[5pt] &= -\sin^2\!x + \cos^2\!x\,\textrm{.} \end{align}

Using the formula for double angles, the answer can be simplified to \displaystyle \cos 2x\,.

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