Question

Find the derivative of ${{\left( 3x+5 \right)}^{4}}\times \sqrt{{{x}^{2}}-1}$

Answer 1

Hint: We solve this question by dividing the given function into two parts $f\left( x \right)={{\left( 3x+5 \right)}^{4}}$ and $g\left( x \right)=\sqrt{{{x}^{2}}-1}$. Then we use the formula for product rule to find the derivative of given function, $\dfrac{d}{dx}\left( f\left( x \right)g\left( x \right) \right)={f}'\left( x \right)g\left( x \right)+f\left( x \right){g}'\left( x \right)$. Then we find the derivatives of both functions using the formula for differentiation $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n\times {{x}^{n-1}}$ and the formula for chain rule, $\dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)={f}'\left( g\left( x \right) \right)\times {g}'\left( x \right)$. Then after finding those derivatives we substitute them in the formula for product rule to find the required derivative.

Complete step by step answer:
Let us assume that $f\left( x \right)={{\left( 3x+5 \right)}^{4}}$ and $g\left( x \right)=\sqrt{{{x}^{2}}-1}$.
So, we can write our given function as $f\left( x \right)g\left( x \right)$.
So, we need to find the derivative of $f\left( x \right)g\left( x \right)$.
Now let us consider the product rule in differentiation.
$\dfrac{d}{dx}\left( f\left( x \right)g\left( x \right) \right)={f}'\left( x \right)g\left( x \right)+f\left( x \right){g}'\left( x \right)$
So, let us find the values of ${f}'\left( x \right)$ and ${g}'\left( x \right)$.
First let us consider the function $f\left( x \right)={{\left( 3x+5 \right)}^{4}}$.
$\Rightarrow {f}'\left( x \right)=\dfrac{d}{dx}{{\left( 3x+5 \right)}^{4}}$
Now let us consider the formula for differentiation of ${{x}^{n}}$,
$\dfrac{d}{dx}\left( {{x}^{n}} \right)=n\times {{x}^{n-1}}$
Let us also consider the formula for chain rule,
$\dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)={f}'\left( g\left( x \right) \right)\times {g}'\left( x \right)$
Using these formulas, we can write ${f}'\left( x \right)$ as
$\begin{align}
  & \Rightarrow {f}'\left( x \right)=4\times {{\left( 3x+5 \right)}^{4-1}}\times \left( \dfrac{d}{dx}\left( 3x+5 \right) \right) \\
 & \Rightarrow {f}'\left( x \right)=4\times {{\left( 3x+5 \right)}^{3}}\times \left( 3 \right) \\
 & \Rightarrow {f}'\left( x \right)=12{{\left( 3x+5 \right)}^{3}} \\
\end{align}$
Now let us consider the function $g\left( x \right)=\sqrt{{{x}^{2}}-1}$.
Now let us consider the formula for differentiation of ${{x}^{n}}$,
$\dfrac{d}{dx}\left( {{x}^{n}} \right)=n\times {{x}^{n-1}}$
Let us also consider the formula for chain rule,
$\dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)={f}'\left( g\left( x \right) \right)\times {g}'\left( x \right)$
Using these formulas, we can write ${g}'\left( x \right)$ as
$\begin{align}
  & \Rightarrow {g}'\left( x \right)=\dfrac{d}{dx}\left( \sqrt{{{x}^{2}}-1} \right) \\
 & \Rightarrow {g}'\left( x \right)=\dfrac{1}{2}{{\left( {{x}^{2}}-1 \right)}^{\dfrac{1}{2}-1}}\times \dfrac{d}{dx}\left( {{x}^{2}}-1 \right) \\
 & \Rightarrow {g}'\left( x \right)=\dfrac{1}{2}\dfrac{1}{\sqrt{{{x}^{2}}-1}}\times \left( 2x \right) \\
 & \Rightarrow {g}'\left( x \right)=\dfrac{x}{\sqrt{{{x}^{2}}-1}} \\
\end{align}$
Substituting these values of functions in the product rule we get,
$\begin{align}
  & \Rightarrow \dfrac{d}{dx}\left( f\left( x \right)g\left( x \right) \right)={f}'\left( x \right)g\left( x \right)+f\left( x \right){g}'\left( x \right) \\
 & \Rightarrow \dfrac{d}{dx}\left( f\left( x \right)g\left( x \right) \right)=12{{\left( 3x+5 \right)}^{3}}\sqrt{{{x}^{2}}-1}+{{\left( 3x+5 \right)}^{4}}\dfrac{x}{\sqrt{{{x}^{2}}-1}} \\
\end{align}$
Solving it by taking ${{\left( 3x+5 \right)}^{3}}$ common we get,
$\begin{align}
  & \Rightarrow \dfrac{d}{dx}\left( f\left( x \right)g\left( x \right) \right)={{\left( 3x+5 \right)}^{3}}\left( 12\sqrt{{{x}^{2}}-1}+\left( 3x+5 \right)\dfrac{x}{\sqrt{{{x}^{2}}-1}} \right) \\
 & \Rightarrow \dfrac{d}{dx}\left( f\left( x \right)g\left( x \right) \right)={{\left( 3x+5 \right)}^{3}}\left( \dfrac{12{{x}^{2}}-12+3{{x}^{2}}+5x}{\sqrt{{{x}^{2}}-1}} \right) \\
 & \Rightarrow \dfrac{d}{dx}\left( f\left( x \right)g\left( x \right) \right)={{\left( 3x+5 \right)}^{3}}\left( \dfrac{15{{x}^{2}}+5x-12}{\sqrt{{{x}^{2}}-1}} \right) \\
\end{align}$
Hence, we get the derivative of ${{\left( 3x+5 \right)}^{4}}\times \sqrt{{{x}^{2}}-1}$ is ${{\left( 3x+5 \right)}^{3}}\left( \dfrac{15{{x}^{2}}+5x-12}{\sqrt{{{x}^{2}}-1}} \right)$.
Hence, answer is ${{\left( 3x+5 \right)}^{3}}\left( \dfrac{15{{x}^{2}}+5x-12}{\sqrt{{{x}^{2}}-1}} \right)$.

Note:
The common mistake made while solving this problem is one might take the formula for product rule as $\dfrac{d}{dx}\left( f\left( x \right)g\left( x \right) \right)={f}'\left( x \right){g}'\left( x \right)+f\left( x \right)g\left( x \right)$, which is wrong because the actual formula for product rule is $\dfrac{d}{dx}\left( f\left( x \right)g\left( x \right) \right)={f}'\left( x \right)g\left( x \right)+f\left( x \right){g}'\left( x \right)$. One might also confuse the formula $\dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)={f}'\left( g\left( x \right) \right)\times {g}'\left( x \right)$ with the formula for product rule. So, one must carefully observe the formulae they are using.