What is the derivative of x\sin x

rheisf 2021-12-29 Answered
What is the derivative of \(\displaystyle{x}{\sin{{x}}}\)

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Bernard Lacey
Answered 2021-12-30 Author has 97 answers
\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={x}{\cos{{x}}}+{\sin{{x}}}\)
Explanation:
We have:
\(\displaystyle{y}={x}{\sin{{x}}}\)
Which is the product of two functions, and so we apply the Product Rule for Differentiation:
\(\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({u}{v}\right)}={u}{\frac{{{d}{v}}}{{{\left.{d}{x}\right.}}}}+{\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}}{v},\ \text{ or }\ {\left({u}{v}\right)}'={\left({d}{u}\right)}{v}+{u}{\left({d}{v}\right)}\)
I was taught to remember the rule in words; "The first times the derivative of the second plus the derivative of the first times the second ".
So with \(\displaystyle{y}={x}{\sin{{x}}}\)
\[\begin{cases}\text{Let} & u = x & \Rightarrow\frac{du}{dx}=1\\\text{And } & v = \sin x & \Rightarrow\frac{dv}{dx}=\cos x\end{cases}\]
Then:
\(\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({u}{v}\right)}={u}{\frac{{{d}{v}}}{{{\left.{d}{x}\right.}}}}+{\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}}{v}\)
Gives us:
\(\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({x}{\sin{{x}}}\right)}={\left({x}\right)}{\left({\cos{{x}}}\right)}+{\left({1}\right)}{\left({\sin{{x}}}\right)}\)
\(\displaystyle\therefore{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={x}{\cos{{x}}}+{\sin{{x}}}\)
If you are new to Calculus then explicitly substituting u and v can be quite helpful, but with practice these steps can be omitted, and the product rule can be applied as we write out the solution.
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Maricela Alarcon
Answered 2021-12-31 Author has 5026 answers
This is a function which is in the form,
\(\displaystyle{y}={f{{\left({x}\right)}}}{g{{\left({x}\right)}}}\)
It's the product of two functions and so we must make use of the product rule. This is a simple formula which you have to remember:
\(\displaystyle\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}={f}'{\left({x}\right)}{g{{\left({x}\right)}}}+{f{{\left({x}\right)}}}{g}'{\left({x}\right)}.\)
In words: the derivative of first function multiplied by the original second function, plus, the derivative of the second function multiplied by the original first function.
In this question,
\(\displaystyle{f{{\left({x}\right)}}}={x}\)
\(\displaystyle{g{{\left({x}\right)}}}={\sin{{\left({x}\right)}}}\)
so we can find that,
\(\displaystyle{f}'{\left({x}\right)}={1}\)
\(\displaystyle{g}'{\left({x}\right)}={\cos{{\left({x}\right)}}}\)
and by substituting this into the formula for the product rule we get the answer:
\(\displaystyle\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}={\sin{{\left({x}\right)}}}+{x}{\cos{{\left({x}\right)}}}.\)
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Vasquez
Answered 2022-01-08 Author has 9055 answers
Finally found the answer to this difficult question, I'm happy
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