Question

Find the derivative of $\dfrac{\sqrt{x}{{\left( x+4 \right)}^{\dfrac{3}{2}}}}{{{\left( 4x-3 \right)}^{\dfrac{4}{3}}}}$with respect to $x$.

Answer 1

Hint: We use the product rule to define the derivative of product of two or more than two functions as $\dfrac{d}{dx}\left( f(x).g(x) \right)={{f}^{'}}(x).g(x)+{{g}^{'}}(x).f(x)$ and quotient rule to define the derivative of two functions like $\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v{{u}^{'}}-u{{v}^{'}}}{{{v}^{2}}}$.

Complete step-by-step solution -
The given function is$\dfrac{\sqrt{x}{{\left( x+4 \right)}^{\dfrac{3}{2}}}}{{{\left( 4x-3 \right)}^{\dfrac{4}{3}}}}$
Clearly , it is of the type $\dfrac{f(x).g(x)}{h(x)}$where $f(x)=\sqrt{x};g(x)={{\left( x+4 \right)}^{\dfrac{3}{2}}}$and $h(x)={{\left( 4x-3 \right)}^{\dfrac{4}{3}}}$
Now to find its derivative , first we will find the expression for the derivative of the numerator. It will be helpful while applying quotient rules to find the derivative of the entire function .
To find the derivative of the numerator , we will apply the product rule of differentiation . We know , the product rule of differentiation is given as $\dfrac{d}{dx}\left( f(x).g(x) \right)={{f}^{'}}(x).g(x)+{{g}^{'}}(x).f(x)$
So, we will differentiate the numerator with respect to $x$ using the product rule of differentiation .
On differentiating the numerator with respect to $x$, we get ,
$\dfrac{d}{dx}\left( \sqrt{x}{{\left( x+4 \right)}^{\dfrac{3}{2}}} \right)=\dfrac{1}{2\sqrt{x}}{{\left( x+4 \right)}^{\dfrac{3}{2}}}+\dfrac{3}{2}{{\left( x+4 \right)}^{\dfrac{1}{2}}}.{{x}^{\dfrac{1}{2}}}$
$={{\left( x+4 \right)}^{\dfrac{1}{2}}}\left[ \dfrac{x+4}{2\sqrt{x}}+\dfrac{3}{2}\sqrt{x} \right]$
$={{\left( x+4 \right)}^{\dfrac{1}{2}}}\left[ \dfrac{x+4+3x}{2\sqrt{x}} \right]$
$={{\left( x+4 \right)}^{\dfrac{1}{2}}}\left[ \dfrac{4x+4}{2\sqrt{x}} \right]$
$={{\left( x+4 \right)}^{\dfrac{1}{2}}}\left[ \dfrac{2x+2}{\sqrt{x}} \right]$
Now, we will differentiate the entire function with respect to $x$. To differentiate the entire function , we will apply the quotient rule of differentiation. For that , first we must know the quotient rule of differentiation. The quotient rule for differentiation is given as $\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v{{u}^{'}}-u{{v}^{'}}}{{{v}^{2}}}$.
So, on differentiating the function by applying quotient rule for the entire function , we get ,
$\dfrac{d}{dx}\left( \dfrac{\sqrt{x}{{\left( x+4 \right)}^{\dfrac{3}{2}}}}{{{\left( 4x-3 \right)}^{\dfrac{4}{3}}}} \right)=\dfrac{{{\left( 4x-3 \right)}^{\dfrac{4}{3}}}.\dfrac{d}{dx}\left( \sqrt{x}{{\left( x+4 \right)}^{\dfrac{3}{2}}} \right)-\left( \sqrt{x}{{\left( x+4 \right)}^{\dfrac{3}{2}}} \right).\dfrac{d}{dx}{{\left( 4x-3 \right)}^{\dfrac{4}{3}}}}{{{\left( {{\left( 4x-3 \right)}^{\dfrac{4}{3}}} \right)}^{2}}}$
$=\dfrac{{{\left( 4x-3 \right)}^{\dfrac{4}{3}}}.{{\left( x+4 \right)}^{\dfrac{1}{2}}}.\left( \dfrac{2x+2}{\sqrt{x}} \right)-\left( \sqrt{x}{{\left( x+4 \right)}^{\dfrac{3}{2}}} \right).\dfrac{4}{3}{{\left( 4x-3 \right)}^{\dfrac{1}{3}}}.4}{{{\left( 4x-3 \right)}^{\dfrac{8}{3}}}}$
Now, we can take ${{\left( 4x-3 \right)}^{\dfrac{1}{3}}}$ common in the numerator. So , we get ,
$\dfrac{d}{dx}\left( \dfrac{\sqrt{x}{{\left( x+4 \right)}^{\dfrac{3}{2}}}}{{{\left( 4x-3 \right)}^{\dfrac{4}{3}}}} \right)=\dfrac{{{\left( 4x-3 \right)}^{\dfrac{1}{3}}}}{{{\left( 4x-3 \right)}^{\dfrac{8}{3}}}}\left[ \left( 4x-3 \right){{\left( x+4 \right)}^{\dfrac{1}{2}}}.\dfrac{2x+2}{\sqrt{x}}-{{\left( x+4 \right)}^{\dfrac{3}{2}}}.\sqrt{x}.\dfrac{16}{3} \right]$
Again , we can take ${{\left( x+4 \right)}^{\dfrac{1}{2}}}$ common in the numerator. So , we get ,
$\dfrac{d}{dx}\left( \dfrac{\sqrt{x}{{\left( x+4 \right)}^{\dfrac{3}{2}}}}{{{\left( 4x-3 \right)}^{\dfrac{4}{3}}}} \right)=\dfrac{{{\left( x+4 \right)}^{\dfrac{1}{2}}}}{{{\left( 4x-3 \right)}^{\dfrac{7}{3}}}}\left[ \dfrac{3\left( 4x-3 \right)\left( 2x+2 \right)-\left( x+4 \right)\left( x \right).16}{3\sqrt{x}} \right]$
$=\dfrac{{{\left( x+4 \right)}^{\dfrac{1}{2}}}}{{{\left( 4x-3 \right)}^{\dfrac{7}{3}}}}\left[ \dfrac{24{{x}^{2}}+6x-18-16{{x}^{2}}-64x}{3\sqrt{x}} \right]$
$=\dfrac{{{\left( x+4 \right)}^{\dfrac{1}{2}}}}{{{\left( 4x-3 \right)}^{\dfrac{7}{3}}}}\left[ \dfrac{8{{x}^{2}}-58x-18}{3\sqrt{x}} \right]$
Hence , the derivative of the function $\dfrac{\sqrt{x}{{\left( x+4 \right)}^{\dfrac{3}{2}}}}{{{\left( 4x-3 \right)}^{\dfrac{4}{3}}}}$with respect to $x$is given as $\dfrac{{{\left( x+4 \right)}^{\dfrac{1}{2}}}}{{{\left( 4x-3 \right)}^{\dfrac{7}{3}}}}\left[ \dfrac{8{{x}^{2}}-58x-18}{3\sqrt{x}} \right]$ .

Note: While differentiating ${{\left( 4x-3 \right)}^{\dfrac{4}{3}}}$, keep in mind that it is a composite function , i.e of the form$h(x)=p(q(x))$ and its derivative will be $\dfrac{4}{3}{{\left( 4x-3 \right)}^{\dfrac{1}{3}}}.4$and not $\dfrac{4}{3}{{\left( 4x-3 \right)}^{\dfrac{1}{3}}}$. Students usually make such mistakes and end up getting the wrong answer. Such mistakes should be avoided .