Question

If \[{x^m}{y^n} = {\left( {x + y} \right)^{m + n}}\]. Then what is the value of \[\left( {\dfrac{{dy}}{{dx}}} \right)\] at \[x = 1\], and \[y = 2\]?
A. \[\dfrac{1}{2}\]
B. 2
C. \[\dfrac{{2m}}{n}\]
D. \[\dfrac{m}{{2n}}\]

Answer 1

Hint: In the given question, one exponential equation is given. To remove the exponents, take \[\log\] on both sides of the equation. Then differentiate the new equation with respect to \[x\]. By substituting the values \[x = 1\] and \[y = 2\] in the differential equation, we will find the value of \[\left( {\dfrac{{dy}}{{dx}}} \right)\] at \[x = 1\], and \[y = 2\].

Formula Used:
\[\log{\left( a \right)^m} = m\log\left( a \right)\]
\[\log\left( {ab} \right) = \log\left( a \right) + \log\left( b \right)\]
\[\dfrac{d}{{dx}}\left( {\log x} \right) = \dfrac{1}{x}\]
\[\dfrac{d}{{dx}}\left( {\log y} \right) = \dfrac{1}{y}\dfrac{{dy}}{{dx}}\]

Complete step by step solution:
The given equation is \[{x^m}{y^n} = {\left( {x + y} \right)^{m + n}}\].
Let’s take \[\log\] on both sides of the above equation.
\[\log\left( {{x^m}{y^n}} \right) = \log{\left( {x + y} \right)^{m + n}}\]
Apply the formulas \[\log{\left( a \right)^m} = m\log\left( a \right)\] and \[\log\left( {ab} \right) = \log\left( a \right) + \log\left( b \right)\]
\[\log\left( {{x^m}} \right) + \log\left( {{y^n}} \right) = \left( {m + n} \right)\log\left( {x + y} \right)\]
Again apply the formula \[\log{\left( a \right)^m} = m\log\left( a \right)\]
\[ \Rightarrow \]\[m\log\left( x \right) + n\log\left( y \right) = \left( {m + n} \right)\log\left( {x + y} \right)\]

Now differentiate the above equation with respect to \[x\].
\[m\dfrac{d}{{dx}}\left( {\log\left( x \right)} \right) + n\dfrac{d}{{dx}}\left( {\log\left( y \right)} \right) = \left( {m + n} \right)\dfrac{d}{{dx}}\left( {\log\left( {x + y} \right)} \right)\]
Apply the formula \[\dfrac{d}{{dx}}\left( {\logy} \right) = \dfrac{1}{y}\dfrac{{dy}}{{dx}}\] and chain rule.
\[ \Rightarrow \]\[m\left( {\dfrac{1}{x}} \right) + n\left( {\dfrac{1}{y}} \right)\left( {\dfrac{{dy}}{{dx}}} \right) = \left( {m + n} \right)\left( {\dfrac{1}{{x + y}}} \right)\left( {1 + \dfrac{{dy}}{{dx}}} \right)\]
\[ \Rightarrow \]\[\left( {\dfrac{m}{x}} \right) + \left( {\dfrac{n}{y}} \right)\left( {\dfrac{{dy}}{{dx}}} \right) = \left( {\dfrac{{m + n}}{{x + y}}} \right)\left( {1 + \dfrac{{dy}}{{dx}}} \right)\]
Apply distributive property \[a\left( {b + c} \right) = ab + ac\]
\[ \Rightarrow \left( {\dfrac{m}{x}} \right) + \left( {\dfrac{n}{y}} \right)\left( {\dfrac{{dy}}{{dx}}} \right) = \left( {\dfrac{{m + n}}{{x + y}}} \right) + \left( {\dfrac{{m + n}}{{x + y}}} \right)\left( {\dfrac{{dy}}{{dx}}} \right)\]
Combine like terms
\[ \Rightarrow \left( {\dfrac{n}{y}} \right)\left( {\dfrac{{dy}}{{dx}}} \right) - \left( {\dfrac{{m + n}}{{x + y}}} \right)\left( {\dfrac{{dy}}{{dx}}} \right) = \left( {\dfrac{{m + n}}{{x + y}}} \right) - \left( {\dfrac{m}{x}} \right)\]
Simplify the above equation:
\[ \Rightarrow \]\[\left( {\dfrac{n}{y} - \dfrac{{m + n}}{{x + y}}} \right)\left( {\dfrac{{dy}}{{dx}}} \right) = \dfrac{{m + n}}{{x + y}} - \dfrac{m}{x}\]
\[ \Rightarrow \]\[\left( {\dfrac{{n\left( {x + y} \right) - y\left( {m + n} \right)}}{{y\left( {x + y} \right)}}} \right)\left( {\dfrac{{dy}}{{dx}}} \right) = \dfrac{{x\left( {m + n} \right) - m\left( {x + y} \right)}}{{x\left( {x + y} \right)}}\]
\[ \Rightarrow \]\[\left( {\dfrac{{nx + ny - my - ny}}{{y\left( {x + y} \right)}}} \right)\left( {\dfrac{{dy}}{{dx}}} \right) = \dfrac{{mx + nx - mx - my}}{{x\left( {x + y} \right)}}\]
\[ \Rightarrow \]\[\left( {\dfrac{{nx - my}}{{y\left( {x + y} \right)}}} \right)\left( {\dfrac{{dy}}{{dx}}} \right) = \dfrac{{nx - my}}{{x\left( {x + y} \right)}}\]
Multiply both sides by \[\dfrac{{\left( {x + y} \right)}}{{\left( {nx - my} \right)}}\].
\[\left( {\dfrac{1}{y}} \right)\left( {\dfrac{{dy}}{{dx}}} \right) = \dfrac{1}{x}\]
\[ \Rightarrow \]\[\left( {\dfrac{{dy}}{{dx}}} \right) = \dfrac{y}{x}\]
Now substitute \[x = 1\] and \[y = 2\] in the above differential equation.
\[\left( {\dfrac{{dy}}{{dx}}} \right) = \dfrac{2}{1}\]
\[ \Rightarrow \]\[\left( {\dfrac{{dy}}{{dx}}} \right) = 2\]

Hence the correct option is option B.
Note: Students are often confused with the formula of \[\log\left( {{x^m}{y^n}} \right)\] whether \[\log\left( {{x^m}} \right) + \log\left( {{y^n}} \right)\] or \[\log\left( {{x^m}} \right) \cdot \log\left( {{y^n}} \right)\]. But the correct formula is \[\log\left( {{x^m}{y^n}} \right) = \log\left( {{x^m}} \right) + \log\left( {{y^n}} \right)\].