Question

How to differentiate the fraction $y = \dfrac{{{{\left( {4x + 1} \right)}^2}}}{{\sqrt {4x + 1} }}$ ?

Answer 1

Here in this question, we have to differentiate the given function. A derivate of an independent value is called differentiation. There are five important rules of differentiation namely power, sum, product, quotient and chain. In order to solve this question, we will make use of some of these rules.

Formula used:
$\sqrt[n]{{{a^x}}} = {a^{\dfrac{x}{n}}}$
$\dfrac{1}{{{b^n}}} = {b^{ - n}}$
${a^m}{a^n} = {a^{m + n}}$
Chain rule: $\dfrac{d}{{dx}}\left[ {f\left( {g\left( x \right)} \right)} \right] = f'\left( {g\left( x \right)} \right)g'\left( x \right)$
Power rule: $\dfrac{d}{{du}}\left[ {{u^n}} \right] = n{u^{n - 1}}$

Complete step by step answer:
The given function is $y = \dfrac{{{{\left( {4x + 1} \right)}^2}}}{{\sqrt {4x + 1} }}$
We will start differentiating this function by first rewriting the denominator $\sqrt {4x + 1}$ by using $\sqrt[n]{{{a^x}}} = {a^{\dfrac{x}{n}}}$ , we get, $\sqrt {4x + 1}$ as ${\left( {4x + 1} \right)^{\dfrac{1}{2}}}$ .
$\Rightarrow \dfrac{d}{{dx}}\left[ {\dfrac{{{{\left( {4x + 1} \right)}^2}}}{{{{\left( {4x + 1} \right)}^{\dfrac{1}{2}}}}}} \right]$
Now, we move ${\left( {4x + 1} \right)^{\dfrac{1}{2}}}$ from denominator to the numerator using the negative exponent rule i.e., $\dfrac{1}{{{b^n}}} = {b^{ - n}}$ and we get,
$\Rightarrow \dfrac{d}{{dx}}\left[ {{{\left( {4x + 1} \right)}^2}{{\left( {4x + 1} \right)}^{ - \dfrac{1}{2}}}} \right]$
Multiply ${\left( {4x + 1} \right)^2}$ with ${\left( {4x + 1} \right)^{ - \dfrac{1}{2}}}$ by adding the exponents. For doing so we will use the power rule i.e., ${a^m}{a^n} = {a^{m + n}}$ to combine the exponents.
$\Rightarrow \dfrac{d}{{dx}}\left[ {{{\left( {4x + 1} \right)}^{2 - \dfrac{1}{2}}}} \right] \\\ \Rightarrow \dfrac{d}{{dx}}\left[ {{{\left( {4x + 1} \right)}^{\dfrac{{4 - 1}}{2}}}} \right] \\\ \Rightarrow \dfrac{d}{{dx}}\left[ {{{\left( {4x + 1} \right)}^{\dfrac{3}{2}}}} \right] \\\$
Now, we will differentiate the obtained function using the chain rule, the chain rule states that $\dfrac{d}{{dx}}\left[ {f\left( {g\left( x \right)} \right)} \right] = f'\left( {g\left( x \right)} \right)g'\left( x \right)$ . Here, $f\left( x \right) = {x^{\dfrac{3}{2}}}$ and $g\left( x \right) = 4x + 1$ . In order to apply the chain rule, we set $u$ as $4x + 1$ and get,
$\Rightarrow \dfrac{d}{{du}}\left[ {{u^{\dfrac{3}{2}}}} \right]\dfrac{d}{{dx}}\left[ {4x + 1} \right]$
Now, we will further differentiate the function using power rule which states that $\dfrac{d}{{du}}\left[ {{u^n}} \right] = n{u^{n - 1}}$ .
$\Rightarrow \dfrac{3}{2}{u^{\dfrac{3}{2} - 1}}\dfrac{d}{{dx}}\left[ {4x + 1} \right] \\\ \Rightarrow \dfrac{3}{2}{\left( {4x + 1} \right)^{\dfrac{3}{2} - 1}}\dfrac{d}{{dx}}\left[ {4x + 1} \right] \\\ \Rightarrow \dfrac{3}{2}{\left( {4x + 1} \right)^{\dfrac{1}{2}}}\dfrac{d}{{dx}}\left[ {4x + 1} \right] \\\ \Rightarrow \dfrac{{3{{\left( {4x + 1} \right)}^{\dfrac{1}{2}}}}}{2}\dfrac{d}{{dx}}\left[ {4x + 1} \right] \\\$
Now, using the sum rule we will find the derivative of $4x + 1$ .
$\Rightarrow \dfrac{{3{{\left( {4x + 1} \right)}^{\dfrac{1}{2}}}}}{2}\left( {\dfrac{d}{{dx}}\left[ {4x} \right] + \dfrac{d}{{dx}}\left[ 1 \right]} \right) \\\ \Rightarrow \dfrac{{3{{\left( {4x + 1} \right)}^{\dfrac{1}{2}}}}}{2}\left( {4\dfrac{d}{{dx}}\left[ x \right] + \dfrac{d}{{dx}}\left[ 1 \right]} \right) \\\ \Rightarrow \dfrac{{3{{\left( {4x + 1} \right)}^{\dfrac{1}{2}}}}}{2}\left( {4\left( 1 \right) + 0} \right) \\\ \Rightarrow \dfrac{{3{{\left( {4x + 1} \right)}^{\dfrac{1}{2}}}}}{2}\left( 4 \right) \\\ \Rightarrow \dfrac{{3{{\left( {4x + 1} \right)}^{\dfrac{1}{2}}} \times 4}}{2} \\\ \Rightarrow \dfrac{{12{{\left( {4x + 1} \right)}^{\dfrac{1}{2}}}}}{2} \\\$
$\Rightarrow 6{\left( {4x + 1} \right)^{\dfrac{1}{2}}}$
Therefore, the derivative of the function $y = \dfrac{{{{\left( {4x + 1} \right)}^2}}}{{\sqrt {4x + 1} }}$ is $6{\left( {4x + 1} \right)^{\dfrac{1}{2}}}$ .

Note: Here in this question, we made use of chain rule, sum rule, product rule, power rule and negative exponent rule to find the derivative of the given function. Students should know when to use these rules and when not to. The common errors which can happen while differentiating this function can be calculation and multiplication errors.