Find the derivative of the Function: P(x) = (x - 3cosx)(x + 3cosx)

Find the derivative of the Function: P(x) = (x - 3cosx)(x + 3cosx)

Question
Derivatives
asked 2021-01-15
Find the derivative of the Function: \(\displaystyle{P}{\left({x}\right)}={\left({x}-{3}{\cos{{x}}}\right)}{\left({x}+{3}{\cos{{x}}}\right)}\)

Answers (1)

2021-01-16
\(\displaystyle{P}{\left({x}\right)}={\left({x}-{3}{\cos{{x}}}\right)}{\left({x}+{3}{\cos{{x}}}\right)}={x}^{{2}}-{9}{{\cos}^{{2}}{\left({x}\right)}}\)
\(\displaystyle\frac{{{d}{p}}}{{{\left.{d}{x}\right.}}}={2}{x}-{9}{d}{\left({{\cos}^{{2}}{x}}\right)}={2}{x}-{9}\cdot{2}{\cos{{x}}}\cdot{d}{\left({\cos{{x}}}\right)}=\)
\(\displaystyle={2}{x}-{9}\cdot{2}{\cos{{x}}}{\left(-{\sin{{x}}}\right)}\)
\(\displaystyle={2}{x}+{9}\cdot{2}{\sin{{x}}}{\cos{{x}}}\)
\(\displaystyle={2}{x}+{9}{\sin{{\left({2}{x}\right)}}}\)
you can use the formula
\(\displaystyle{\left({a}-{b}\right)}{\left({a}+{b}\right)}={a}^{{2}}-{b}^{{2}}\)
\(\displaystyle{2}{\sin{{x}}}{\cos{{x}}}={\sin{{\left({2}{x}\right)}}}\)
or you can use the product rule
\(\displaystyle\frac{{{d}{p}}}{{{\left.{d}{x}\right.}}}={\left({d}{\left({x}-{3}{\cos{{x}}}\right)}\right)}{\left({x}+{3}{\cos{{x}}}\right)}+{\left({x}-{3}{\cos{{x}}}\right)}{\left({d}{\left({x}+{3}{\cos{{x}}}\right)}\right)}=\)
\(\displaystyle{\left({1}+{3}{\sin{{x}}}\right)}{\left({x}+{3}{\cos{{x}}}\right)}+{\left({x}-{3}{\cos{{x}}}\right)}{\left({1}-{3}{\sin{{x}}}\right)}=\)
\(\displaystyle{x}+{3}{\cos{{x}}}+{3}{x}{\sin{{x}}}+{9}{\sin{{x}}}{\cos{{x}}}+{x}-{3}{\cos{{x}}}-{3}{x}{\sin{{x}}}+{9}{\sin{{x}}}{\cos{{x}}}=\)
\(\displaystyle{2}{x}+{18}{\sin{{x}}}{\cos{{x}}}={2}{x}+{9}\cdot{2}{\sin{{x}}}{\cos{{x}}}={2}{x}+{9}{\sin{{\left({2}{x}\right)}}}\)
0

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