Question

How do you differentiate $g\left( z \right)={{z}^{2}}\cos \left( 3z \right){{e}^{2+Z}}$ using the product rule?

Answer 1

Hint: From the given question we have to find the differentiation of $g\left( z \right)={{z}^{2}}\cos \left( 3z \right){{e}^{2+Z}}$ by using product rule. Product rule or UV rule means when we differentiate any function in the form UV then we will have to differentiate like \[D\left( UV \right)=D\left( V \right)\times U+D\left( U \right)\times V\]. But here in the given function there are three terms so that the product rule will be $D\left( UVW \right)=D\left( W \right)\times U\times V+D\left( V \right)\times U\times W+D\left( U \right)\times V\times W$. And we have to use the derivative of the two-composition function or chain rule that is $\dfrac{d}{dx}\left( f\left[ g\left( x \right) \right] \right)=f'\left[ g\left( x \right) \right]\times g'\left( x \right)$. By this we can find the differentiation of the given function.

Complete step-by-step solution:
From the question given we have to find the differentiation of a given function by using product rule, the given function is
$\Rightarrow g\left( z \right)={{z}^{2}}\cos \left( 3z \right){{e}^{2+Z}}$
Here we have to solve this by using product rule.
As we know that the product rule or UV rule means when we differentiate any function in the form UV then we will have to differentiate like,
\[\Rightarrow D\left( UV \right)=D\left( V \right)\times U+D\left( U \right)\times V\]
 But here in the given function there are three terms so that the product rule will be,
$\Rightarrow D\left( UVW \right)=D\left( W \right)\times U\times V+D\left( V \right)\times U\times W+D\left( U \right)\times V\times W$
Now we will write the differentiation of each term separately,
First differentiation of ${{z}^{2}}$
$\Rightarrow \dfrac{d}{dz}\left( {{z}^{2}} \right)=2z$
Now differentiation of $\cos \left( 3z \right)$
Here we have to use the chain rule or the derivative of two composition function, that is
$\Rightarrow \dfrac{d}{dx}\left( f\left[ g\left( x \right) \right] \right)=f'\left[ g\left( x \right) \right]\times g'\left( x \right)$
By this the differentiation of $\cos \left( 3z \right)$
$\Rightarrow \dfrac{d}{dz}\left( \cos \left( 3z \right) \right)=-3\sin \left( 3z \right)$
Now the differentiation of ${{e}^{2+Z}}$
$\Rightarrow \dfrac{d}{dz}\left( {{e}^{2+Z}} \right)={{e}^{2+Z}}\left( 0+1 \right)={{e}^{2+Z}}$
Now substituting the all the derivative in the product rule, we will get,
$\Rightarrow D\left( UVW \right)=D\left( W \right)\times U\times V+D\left( V \right)\times U\times W+D\left( U \right)\times V\times W$
$\Rightarrow \dfrac{d}{dz}\left( g\left( z \right) \right)=\dfrac{d}{dz}\left( {{z}^{2}}\cos \left( 3z \right){{e}^{2+Z}} \right)$
$\Rightarrow g'\left( z \right)=2z\times \cos \left( 3z \right)\times {{e}^{2+Z}}+\left( -3\sin \left( 3z \right) \right)\times {{z}^{2}}\times {{e}^{2+Z}}+{{z}^{2}}\times \cos \left( 3z \right)\times {{e}^{2+Z}}$
By simplifying further, we will get,
$\Rightarrow g'\left( z \right)=z{{e}^{2+Z}}\left[ 2\cos \left( 3z \right)\left( 1+z \right)-3z\sin \left( 3z \right) \right]$
Therefore, the required answer is shown above by using the product rule.

Note: Students should know the basic formulas of differentiation and the rules like chain rule, product or UV rule etc. students should also know the formulas like,
$\begin{align}
  & \Rightarrow \dfrac{d\left( {{x}^{n}} \right)}{dx}=n\times {{x}^{n-1}} \\
 & \Rightarrow \dfrac{d}{dx}\left( \sin x \right)=\cos x \\
 & \Rightarrow \dfrac{d}{dx}\left( \cos x \right)=-\sin x \\
 & \Rightarrow \dfrac{d}{dx}\left( \tan x \right)={{\sec }^{2}}x \\
\end{align}$