Question

If \[f\left( x \right) = {\left[ {\dfrac{{\left( {a + x} \right)}}{{\left( {b + x} \right)}}} \right]^{a + b + 2x}}\] , then find the value of \[f'\left( 0 \right)\].
A. \[2log\dfrac{a}{b} + \dfrac{{{b^2} - {a^2}}}{{ab}}\]
B. \[{\left( {\dfrac{a}{b}} \right)^{a + b}}\left( {2log\dfrac{a}{b} + \dfrac{{{b^2} - {a^2}}}{{ab}}} \right)\]
C. \[{\left( {\dfrac{a}{b}} \right)^{a + b}}\left( {\dfrac{{{b^2} - {a^2}}}{{ab}}} \right)\]
D. None of these

Answer 1

Hint: In the question, the given equation is an exponential equation. To remove the exponent from the equation, we will apply the \[\log \] on both sides. Then we will find the derivative of the new logarithmic equation with respect to \[x\]. By substituting \[x = 0\] in the differential equation, we will find the value of \[f'\left( 0 \right)\].


Formula Used:\[lo{g_m}{\left( a \right)^n} = nlo{g_m}\left( a \right)\]
\[lo{g_m}\left( {\dfrac{a}{b}} \right) = lo{g_m}\left( a \right) - lo{g_m}\left( b \right)\]
\[\dfrac{d}{{dx}}\left( {log{x^n}} \right) = \dfrac{1}{{{x^n}}}\dfrac{d}{{dx}}\left( {{x^n}} \right) = \dfrac{n}{x}\]
Chain rule: \[\dfrac{d}{{dx}}\left( {uv} \right) = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}\]
\[\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}\]


Complete step-by-step solution:The given equation is \[f\left( x \right) = {\left[ {\dfrac{{\left( {a + x} \right)}}{{\left( {b + x} \right)}}} \right]^{a + b + 2x}}\] ……..(1)
Let’s take the \[\log \] on both sides of the equation.
\[logf\left( x \right) = \left( {a + b + 2x} \right)log\left[ {\dfrac{{\left( {a + x} \right)}}{{\left( {b + x} \right)}}} \right]\]
Simplify the above equation
\[logf\left( x \right) = \left( {a + b + 2x} \right)\left[ {log\left( {a + x} \right) - log\left( {b + x} \right)} \right]\]

Let’s calculate the derivative of the above equation with respect to \[x\].
Now we will apply \[\dfrac{d}{{dx}}\left( {log{x^n}} \right) = \dfrac{1}{{{x^n}}}\dfrac{d}{{dx}}\left( {{x^n}} \right) = \dfrac{n}{x}\] on left-hand side and chain rule on right-hand side.
\[\dfrac{1}{{f\left( x \right)}}f'\left( x \right) = \left( {a + b + 2x} \right)\left[ {\dfrac{1}{{\left( {a + x} \right)}} - \dfrac{1}{{\left( {b + x} \right)}}} \right] + 2\left[ {log\left( {a + x} \right) - log\left( {b + x} \right)} \right]\] \ …..(2)

Now substitute \[x = 0\] in the equation (1).
\[f\left( 0 \right) = {\left[ {\dfrac{{\left( {a + 0} \right)}}{{\left( {b + 0} \right)}}} \right]^{a + b + 2\left( 0 \right)}}\]
\[ \Rightarrow \]\[f\left( 0 \right) = {\left( {\dfrac{a}{b}} \right)^{a + b}}\]
Substitute \[x = 0\] in the equation (2).
       \[\dfrac{{f'\left( 0 \right)}}{{f\left( 0 \right)}} = \left( {a + b + 2\left( 0 \right)} \right)\left[ {\dfrac{1}{{\left( {a + 0} \right)}} - \dfrac{1}{{\left( {b + 0} \right)}}} \right] + 2\left[ {log\left( {a + 0} \right) - log\left( {b + 0} \right)} \right]\]
Simplify the above equation
\[\dfrac{{f'\left( 0 \right)}}{{f\left( 0 \right)}} = \left( {a + b} \right)\left[ {\dfrac{1}{a} - \dfrac{1}{b}} \right] + 2\left[ {log\left( a \right) - log\left( b \right)} \right]\]
Now we will apply the formulas \[\dfrac{a}{b} - \dfrac{c}{d} = \dfrac{{ad - bc}}{{bd}}\] and \[lo{g_m}\left( a \right) - lo{g_m}\left( b \right) = lo{g_m}\left( {\dfrac{a}{b}} \right)\].
\[\dfrac{{f'\left( 0 \right)}}{{f\left( 0 \right)}} = \left( {a + b} \right)\left[ {\dfrac{{b - a}}{{ab}}} \right] + 2log\left( {\dfrac{a}{b}} \right)\]
Now we will apply the formula \[\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}\].
\[\dfrac{{f'\left( 0 \right)}}{{f\left( 0 \right)}} = \dfrac{{{b^2} - {a^2}}}{{ab}} + 2log\left( {\dfrac{a}{b}} \right)\]
Multiply both sides of the above equation by \[f\left( 0 \right)\].
\[f'\left( 0 \right) = f\left( 0 \right)\left( {\dfrac{{{b^2} - {a^2}}}{{ab}} + 2log\left( {\dfrac{a}{b}} \right)} \right)\]
\[ \Rightarrow \]\[f'\left( 0 \right) = {\left( {\dfrac{a}{b}} \right)^{a + b}}\left( {\dfrac{{{b^2} - {a^2}}}{{ab}} + 2log\left( {\dfrac{a}{b}} \right)} \right)\] [since \[f\left( 0 \right) = {\left( {\dfrac{a}{b}} \right)^{a + b}}\]]

Hence the correct option is option B.

Note: Students are often confused with the formula \[\dfrac{d}{{dx}}\left( {log{x^n}} \right) = \dfrac{1}{{{x^n}}}\] and \[\dfrac{d}{{dx}}\left( {log{x^n}} \right) = \dfrac{1}{{{x^n}}}\dfrac{d}{{dx}}\left( {{x^n}} \right) = \dfrac{n}{x}\]. But the correct formula is \[\dfrac{d}{{dx}}\left( {log{x^n}} \right) = \dfrac{n}{x}\]. Because we also have to multiply the derivative of \[log{x^n}\]by the derivative of the term \[{x^n}\].