\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={x}{\cos{{x}}}+{\sin{{x}}}\)

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.

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.