# What is the derivative of x\sin x

What is the derivative of $$\displaystyle{x}{\sin{{x}}}$$

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Bernard Lacey
$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={x}{\cos{{x}}}+{\sin{{x}}}$$
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
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:
$$\displaystyle\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}={\sin{{\left({x}\right)}}}+{x}{\cos{{\left({x}\right)}}}.$$
Vasquez
Finally found the answer to this difficult question, I'm happy