Question

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

Answer 1

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,

& \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{aligned}$$ Hence, the above relation is our answer. **Note:** One may note that whenever we are asked to differentiate a product of two or more functions we apply the product rule. You must remember all the basic rules and formulas of differentiation like: - the product rule, chain rule, $$\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.