Question

If $y = f\left( {\dfrac{{\left( {2x + 3} \right)}}{{\left( {3 - 2x} \right)}}} \right)$ and $f'\left( x \right) = \sin \left( {\log x} \right)$, then $\dfrac{{dy}}{{dx}}$ is equal to
A) $\dfrac{{\sin \left( {\log x} \right)}}{{x\log x}}$
B) $\left( {\dfrac{{12}}{{{{\left( {3 - 2x} \right)}^2}}}} \right)\left[ {\sin \log \dfrac{{\left( {2x + 3} \right)}}{{\left( {3 - 2x} \right)}}} \right]$
C) $\dfrac{{\cos \left( {\log x} \right)}}{{x\log x}}$
D) None of these

Answer 1

In order to find the value of $\dfrac{{dy}}{{dx}}$, differentiate the given function of ‘y’ with respect to ‘x’, and solve it further using the differentiation rule of quotient i.e $\dfrac{{d\left( {\dfrac{u}{v}} \right)}}{{dx}} = \dfrac{{v\dfrac{{du}}{{dx}} - u\dfrac{{dv}}{{dx}}}}{{{v^2}}}$ and other methods. If needed, substitute the value of $f'\left( x \right) = \sin \left( {\log x} \right)$.

Formula used:
$\dfrac{{d\left( {\dfrac{u}{v}} \right)}}{{dx}} = \dfrac{{v\dfrac{{du}}{{dx}} - u\dfrac{{dv}}{{dx}}}}{{{v^2}}}$

Complete step by step answer:
We are given one functions $y = f\left( {\dfrac{{\left( {2x + 3} \right)}}{{\left( {3 - 2x} \right)}}} \right)$ ……….(1)
And, a first order derivative of a function:
$f'\left( x \right) = \sin \left( {\log x} \right)$ …..(2)
On Differentiating equation 1, we get:
$\dfrac{{dy}}{{dx}} = \dfrac{{d\left( {f\left( {\dfrac{{\left( {2x + 3} \right)}}{{\left( {3 - 2x} \right)}}} \right)} \right)}}{{dx}}$
Since, from differentiation rules, we know that $\dfrac{{d\left( {f\left( n \right)} \right)}}{{dx}}$ can be written as $\dfrac{{d\left( {f\left( n \right)} \right)}}{{dx}} = f'\left( n \right)\dfrac{{dn}}{{dx}}$.
So, comparing our equation $\dfrac{{dy}}{{dx}} = \dfrac{{d\left( {f\left( {\dfrac{{\left( {2x + 3} \right)}}{{\left( {3 - 2x} \right)}}} \right)} \right)}}{{dx}}$ with the above equation $\dfrac{{d\left( {f\left( n \right)} \right)}}{{dx}} = f'\left( n \right)\dfrac{{dn}}{{dx}}$, we get:
$\dfrac{{d\left( {f\left( {\dfrac{{\left( {2x + 3} \right)}}{{\left( {3 - 2x} \right)}}} \right)} \right)}}{{dx}} = f'\left( {\dfrac{{\left( {2x + 3} \right)}}{{\left( {3 - 2x} \right)}}} \right)\dfrac{{d\left( {\dfrac{{\left( {2x + 3} \right)}}{{\left( {3 - 2x} \right)}}} \right)}}{{dx}}$ ……(3)
From the differentiation rule for division, also known as quotient rule, we know that:
$\dfrac{{d\left( {\dfrac{u}{v}} \right)}}{{dx}} = \dfrac{{v\dfrac{{du}}{{dx}} - u\dfrac{{dv}}{{dx}}}}{{{v^2}}}$ ……..(4)
Comparing the above value with $\dfrac{{d\left( {\dfrac{{\left( {2x + 3} \right)}}{{\left( {3 - 2x} \right)}}} \right)}}{{dx}}$, we get:
$u = 2x + 3$
$v = 3 - 2x$
Differentiating u w.r.to x, we get:
$\dfrac{{du}}{{dx}} = \dfrac{{d\left( {2x + 3} \right)}}{{dx}} = 2\dfrac{{dx}}{{dx}} + \dfrac{{d3}}{{dx}}$
$\Rightarrow \dfrac{{du}}{{dx}} = 2$ ….(5)
Differentiating v w.r.to x, we get:
$\dfrac{{dv}}{{dx}} = \dfrac{{d\left( {3 - 2x} \right)}}{{dx}} = \dfrac{{d3}}{{dx}} - 2\dfrac{{dx}}{{dx}}$
$\Rightarrow \dfrac{{dv}}{{dx}} = - 2$ ….(6)
Comparing the value $\dfrac{{d\left( {\dfrac{u}{v}} \right)}}{{dx}} = \dfrac{{v\dfrac{{du}}{{dx}} - u\dfrac{{dv}}{{dx}}}}{{{v^2}}}$ with $\dfrac{{d\left( {\dfrac{{\left( {2x + 3} \right)}}{{\left( {3 - 2x} \right)}}} \right)}}{{dx}}$, we get:
$\dfrac{{d\left( {\dfrac{{\left( {2x + 3} \right)}}{{\left( {3 - 2x} \right)}}} \right)}}{{dx}} = \dfrac{{\left( {3 - 2x} \right)\dfrac{{d\left( {2x + 3} \right)}}{{dx}} - \left( {2x + 3} \right)\dfrac{{d\left( {3 - 2x} \right)}}{{dx}}}}{{{{\left( {3 - 2x} \right)}^2}}}$
Substituting the value of $\dfrac{{dv}}{{dx}}$ and $\dfrac{{du}}{{dx}}$ from the equation (5) and (6) in the above equation, we get:
$\Rightarrow \dfrac{{d\left( {\dfrac{{\left( {2x + 3} \right)}}{{\left( {3 - 2x} \right)}}} \right)}}{{dx}} = \dfrac{{\left( {3 - 2x} \right)2 - \left( {2x + 3} \right)\left( { - 2} \right)}}{{{{\left( {3 - 2x} \right)}^2}}}$
Solving it further, we get:
$\Rightarrow \dfrac{{d\left( {\dfrac{{\left( {2x + 3} \right)}}{{\left( {3 - 2x} \right)}}} \right)}}{{dx}} = \dfrac{{6 - 4x + 4x + 6}}{{{{\left( {3 - 2x} \right)}^2}}} = \dfrac{{12}}{{{{\left( {3 - 2x} \right)}^2}}}$ …..(7)
Since, we are given that $f'\left( x \right) = \sin \left( {\log x} \right)$, substituting the value of $x$ as $\dfrac{{\left( {2x + 3} \right)}}{{\left( {3 - 2x} \right)}}$, we get:
$f'\left( {\dfrac{{\left( {2x + 3} \right)}}{{\left( {3 - 2x} \right)}}} \right) = \sin \left( {\log \dfrac{{\left( {2x + 3} \right)}}{{\left( {3 - 2x} \right)}}} \right)$ ………(8)
Substituting the value of equation (7) and equation (8) in the equation (3), we get:
$\dfrac{{d\left( {f\left( {\dfrac{{\left( {2x + 3} \right)}}{{\left( {3 - 2x} \right)}}} \right)} \right)}}{{dx}} = f'\left( {\dfrac{{\left( {2x + 3} \right)}}{{\left( {3 - 2x} \right)}}} \right)\dfrac{{d\left( {\dfrac{{\left( {2x + 3} \right)}}{{\left( {3 - 2x} \right)}}} \right)}}{{dx}}$
$\Rightarrow \dfrac{{d\left( {f\left( {\dfrac{{\left( {2x + 3} \right)}}{{\left( {3 - 2x} \right)}}} \right)} \right)}}{{dx}} = \dfrac{{12}}{{{{\left( {3 - 2x} \right)}^2}}}\left[ {\sin \left( {\log \dfrac{{\left( {2x + 3} \right)}}{{\left( {3 - 2x} \right)}}} \right)} \right]$
That is,
$\Rightarrow \dfrac{{d\left( y \right)}}{{dx}} = \dfrac{{12}}{{{{\left( {3 - 2x} \right)}^2}}}\left[ {\sin \left( {\log \dfrac{{\left( {2x + 3} \right)}}{{\left( {3 - 2x} \right)}}} \right)} \right]$
Hence, Option (B) is correct.

Note:
We can also further solve the equation for any point by substituting the value of x as 1, 2 or etc, in the above equation and can also expand the logarithmic term if needed using the formula $\log \dfrac{a}{b} = \log a - \log b$.
Since, there are one function and one value inside the function in the value $\dfrac{{d\left( {f\left( n \right)} \right)}}{{dx}} = f'\left( n \right)\dfrac{{dn}}{{dx}}$, so we differentiated the whole function then the internal value, according to the rules of differentiation.