 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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$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={x}{\cos{{x}}}+{\sin{{x}}}$$
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.
Not exactly what you’re looking for? 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