Courses
Courses for Kids
Free study material
Free LIVE classes
More # Find the value of $\dfrac{{dy}}{{dx}}$ if ${x^p}{y^q} = {\left( {x + y} \right)^{p + q}}$.(A) $- \dfrac{x}{y}$ (B) $\dfrac{x}{y}$ (C) $- \dfrac{y}{x}$ (D) $\dfrac{y}{x}$

Last updated date: 16th Mar 2023
Total views: 303.6k
Views today: 7.83k Verified
303.6k+ views
Hint: Take logarithm on both sides of the equation. Use formulae $\log mn = \log m + \log n$ and $\log {m^n} = n\log m$ and then differentiate both sides with respect to $x$.

According to question, the given equation is:
$\Rightarrow {x^p}{y^q} = {\left( {x + y} \right)^{p + q}}$.
Taking logarithm on both sides of this equation, we’ll get:
$\Rightarrow \log \left( {{x^p}{y^q}} \right) = \log {\left( {x + y} \right)^{p + q}},$
We know that $\log mn = \log m + \log n$ and $\log {m^n} = n\log m$, applying these formulae, we’ll get:
$\Rightarrow \log {x^p} + \log {y^q} = \left( {p + q} \right)\log \left( {x + y} \right), \\ \Rightarrow p\log x + q\log y = \left( {p + q} \right)\log \left( {x + y} \right) \\$
Now, differentiating both sides with respect to $x$, we’ll get:
$\Rightarrow \dfrac{d}{{dx}}\left( {p\log x + q\log y} \right) = \dfrac{d}{{dx}}\left[ {\left( {p + q} \right)\log \left( {x + y} \right)} \right], \\ \Rightarrow p\dfrac{d}{{dx}}\log x + q\dfrac{d}{{dx}}\log y = \left( {p + q} \right)\dfrac{d}{{dx}}\log \left( {x + y} \right) .....(i) \\$
We know that $\dfrac{d}{{dx}}\log x = \dfrac{1}{x}$. And for $\dfrac{d}{{dx}}\left( {x + y} \right)$, we will use chain rule of differentiation which is:
$\Rightarrow \dfrac{d}{{dx}}f\left( {g\left( x \right)} \right) = f'\left( {g\left( x \right)} \right) \times \dfrac{d}{{dx}}g\left( x \right)$.
Using these formulae in equation $(i)$, we’ll get:
$\Rightarrow \dfrac{p}{x} + \dfrac{q}{y}\dfrac{{dy}}{{dx}} = \left( {p + q} \right)\left[ {\dfrac{1}{{\left( {x + y} \right)}} \times \dfrac{d}{{dx}}\left( {x + y} \right)} \right], \\ \Rightarrow \dfrac{p}{x} + \dfrac{q}{y}\dfrac{{dy}}{{dx}} = \dfrac{{\left( {p + q} \right)}}{{\left( {x + y} \right)}} \times \left( {1 + \dfrac{{dy}}{{dx}}} \right), \\$
$\Rightarrow \dfrac{p}{x} + \dfrac{q}{y}\dfrac{{dy}}{{dx}} = \dfrac{{\left( {p + q} \right)}}{{\left( {x + y} \right)}} + \dfrac{{\left( {p + q} \right)}}{{\left( {x + y} \right)}}\dfrac{{dy}}{{dx}},$
Now separating the terms having $\dfrac{{dy}}{{dx}}$ on one side, we’ll get:
$\Rightarrow \dfrac{p}{x} - \dfrac{{\left( {p + q} \right)}}{{\left( {x + y} \right)}} = \left[ {\dfrac{{\left( {p + q} \right)}}{{\left( {x + y} \right)}} - \dfrac{q}{y}} \right]\dfrac{{dy}}{{dx}}, \\ \Rightarrow \dfrac{{px + py - px - qx}}{{x\left( {x + y} \right)}} = \dfrac{{\left( {py + qy - qx - qy} \right)}}{{y\left( {x + y} \right)}}\dfrac{{dy}}{{dx}}, \\ \Rightarrow \dfrac{{py - qx}}{x} = \dfrac{{\left( {py - qx} \right)}}{y}\dfrac{{dy}}{{dx}}, \\ \Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = \dfrac{1}{x}, \\ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{y}{x} \\$
Thus, the value of $\dfrac{{dy}}{{dx}}$ is $\dfrac{y}{x}$. (D) is the correct option.

Note: We can also directly differentiate the equation ${x^p}{y^q} = {\left( {x + y} \right)^{p + q}}$ without using logarithm:
$\Rightarrow \dfrac{d}{{dx}}\left( {{x^p}{y^q}} \right) = \dfrac{d}{{dx}}{\left( {x + y} \right)^{p + q}} .....(ii)$
For $\dfrac{d}{{dx}}\left( {{x^p}{y^q}} \right)$, we will use product rule of differentiation which is given as:
$\dfrac{d}{{dx}}\left[ {f\left( x \right)g\left( x \right)} \right] = f\left( x \right)\dfrac{d}{{dx}}g\left( x \right) + g\left( x \right)\dfrac{d}{{dx}}f\left( x \right)$
And for $\dfrac{d}{{dx}}{\left( {x + y} \right)^{p + q}}$, we will use formula $\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$ and then chain rule discussed above.
Using all these results in equation $(ii)$, we’ll get:
$\Rightarrow {x^p}\dfrac{d}{{dx}}{y^q} + {y^q}\dfrac{d}{{dx}}{x^p} = \left( {p + q} \right){\left( {x + y} \right)^{p + q - 1}}\dfrac{d}{{dx}}\left( {x + y} \right)$
When we differentiate the terms and separate them, we’ll get the same result as we have found above.