Question

How do you find the derivative of \[q\left( r \right)={{r}^{3}}\cos r\]?

Answer 1

Hint: Consider q(r) as the product of an algebraic function and a trigonometric function. Now, apply the product rule of differentiation given as: - \[\dfrac{d\left( u\times v \right)}{dr}=u\dfrac{dv}{dr}+v\dfrac{du}{dr}\]. Here, consider, \[u={{r}^{3}}\] and \[v=\cos r\]. Use the formula: - \[\dfrac{d\cos r}{dr}=-\sin r\] to differentiate the trigonometric function and power reduction formula given as: $\dfrac{d\left[ {{r}^{n}} \right]}{dr}=n{{r}^{n-1}}$, where n = 3, to simplify the derivative and get the answer.

Complete step by step solution:
Here, we have been provided with the function \[q\left( r \right)={{r}^{3}}\cos r\] and we are asked to differentiate it. Here we are going to use the product rule of differentiation to get the answer.
\[\because q\left( r \right)={{r}^{3}}\cos r\]
Clearly, we can see that we have q(r) as a function of r. Now, we can assume the given function as the product of an algebraic function ${{r}^{3}}$ and a trigonometric function \[\left( \cos r \right)\]. So, we have,
\[\Rightarrow q\left( r \right)={{r}^{3}}\times \cos r\]
Let us assume ${{r}^{3}}$ as ‘u’ and \[\cos r\] as ‘v’. So, we have,
\[\Rightarrow q\left( r \right)=u\times v\]
Differentiating both the sides with respect to r, we get,
\[\Rightarrow \dfrac{d\left[ q\left( r \right) \right]}{dr}=\dfrac{d\left( u\times v \right)}{dr}\]
Now, applying the product rule of differentiation given as: - \[\dfrac{d\left( u\times v \right)}{dr}=u\dfrac{dv}{dr}+v\dfrac{du}{dr}\], we get,
\[\Rightarrow \dfrac{d\left[ q\left( r \right) \right]}{dr}=\left[ u\dfrac{dv}{dr}+v\dfrac{du}{dr} \right]\]
Substituting the assumed values of u and v, we get,
\[\Rightarrow \dfrac{d\left[ q\left( r \right) \right]}{dr}=\left[ {{r}^{3}}\dfrac{d\cos r}{dr}+\cos r\dfrac{d\left[ {{r}^{3}} \right]}{dr} \right]\]
We know that \[\dfrac{d\cos r}{dr}=-\sin r\], so we have,
\[\Rightarrow \dfrac{d\left[ q\left( r \right) \right]}{dr}=\left[ {{r}^{3}}\left( -\sin r \right)+\cos r\dfrac{d\left[ {{r}^{3}} \right]}{dr} \right]\]
Now, the derivative of the function ${{r}^{3}}$ can be given by using the power reduction formula: $\dfrac{d\left[ {{r}^{n}} \right]}{dr}=n{{r}^{n-1}}$, where n = 3, so we get,
\[\begin{align}
  & \Rightarrow \dfrac{d\left[ q\left( r \right) \right]}{dr}=\left[ -{{r}^{3}}\sin r+\cos r\left( 3\times {{r}^{2}} \right) \right] \\
 & \Rightarrow \dfrac{d\left[ q\left( r \right) \right]}{dr}=\left[ -{{r}^{3}}\sin r+3{{r}^{2}}\cos r \right] \\
 & \Rightarrow \dfrac{d\left[ q\left( r \right) \right]}{dr}={{r}^{2}}\left( 3\cos r-r\sin r \right) \\
\end{align}\]
Hence, the above relation is our answer.

Note: One may note that whenever we are asked to differentiate a product of two or more functions we apply the product rule. You must remember all the basic rules and formulas of differentiation like: - the product rule, chain rule, \[\dfrac{u}{v}\] rule etc. as they are frequently used in both differential and integral calculus. Remember the derivatives of some common functions like: algebraic functions, trigonometric functions, logarithmic functions, exponential functions etc. as we may be asked to find the derivative of the product of any two of these listed functions.