Question

If we have an expression \[{x^m}{y^n} = {(x + y)^{m + n}}\], then ${(\dfrac{{dy}}{{dx}})_{x = 1,y = 2}}$ is equal to
$A)\dfrac{1}{2}$
$B)2$
$C)\dfrac{{2m}}{n}$
$D)\dfrac{m}{{2n}}$

Answer 1

Hint: First, we need to analyze the given information which is in algebraic form.
We can equate the given expression into some form and then we can differentiate using the derivatives of basic functions and applying the chain rule of differentiation.
differentiation, the derivative of $x$ raised to the power is denoted by $\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$ .
We know that the logarithm derivative function can be represented as $\dfrac{d}{{dx}}\log x = \dfrac{1}{x}$
Formula used:
Chain rule of differentiation $\dfrac{d}{{dx}}(f(g(x)) = {f^1}(g(x)) \times {g^1}(x)$

Complete step-by-step solution:
Since from the given that we have, \[{x^m}{y^n} = {(x + y)^{m + n}}\]
Now taking the logarithm function on both sides then we get \[{x^m}{y^n} = {(x + y)^{m + n}} \Rightarrow m\log x + n\log y = (m + n)\log (x + y)\] where the general log can be expressed as $\log {x^m} = m\log x$ (we applied this method in the above equation)
Now taking derivative with respect to x, on both sides we get \[m\log x + n\log y = (m + n)\log (x + y) \Rightarrow \dfrac{{d(m\log x)}}{{dx}} + \dfrac{{d(n\log y)}}{{dx}} = \dfrac{{d((m + n)\log (x + y))}}{{dx}}\]
Since, \[\dfrac{{d(m\log x)}}{{dx}} = \dfrac{m}{x}\] where $\dfrac{d}{{dx}}\log x = \dfrac{1}{x}$
Similarly, \[\dfrac{{d(n\log y)}}{{dx}} = \dfrac{n}{y}(\dfrac{{dy}}{{dx}})\] (because we are derivation with respect to x only, so theY-terms does not differ)
Also, \[\dfrac{{d((m + n)\log (x + y))}}{{dx}} = \dfrac{{m + n}}{{x + y}}(1 + \dfrac{{dy}}{{dx}})\] with the help of chain rule, $\dfrac{d}{{dx}}(f(g(x)) = {f^1}(g(x)) \times {g^1}(x)$
Hence, we get \[\dfrac{{d(m\log x)}}{{dx}} + \dfrac{{d(n\log y)}}{{dx}} = \dfrac{{d((m + n)\log (x + y))}}{{dx}} \Rightarrow \dfrac{m}{x} + \dfrac{n}{y}(\dfrac{{dy}}{{dx}}) = \dfrac{{m + n}}{{x + y}}(1 + \dfrac{{dy}}{{dx}})\]
Now taking the derivative parts on the left side and remaining values on the right side we get \[\dfrac{m}{x} + \dfrac{n}{y}(\dfrac{{dy}}{{dx}}) = \dfrac{{m + n}}{{x + y}}(1 + \dfrac{{dy}}{{dx}}) \Rightarrow \dfrac{n}{y}(\dfrac{{dy}}{{dx}}) - \dfrac{{m + n}}{{x + y}}\dfrac{{dy}}{{dx}} = \dfrac{{m + n}}{{x + y}} - \dfrac{m}{x}\]
Further simplifying we get \[ (\dfrac{{dy}}{{dx}})[\dfrac{n}{y} - \dfrac{{m + n}}{{x + y}}] = \dfrac{{m + n}}{{x + y}} - \dfrac{m}{x}\] (taking out the common values)
Now cross multiplying and solving the equation we have, \[ (\dfrac{{dy}}{{dx}})[\dfrac{{nx + ny - my - ny}}{{y(x + y)}}] = \dfrac{{mx + nx - mx - my}}{{x(x + y)}}\]
\[ \Rightarrow (\dfrac{{dy}}{{dx}})[\dfrac{{nx - my}}{{y(x + y)}}] = \dfrac{{nx - my}}{{x(x + y)}}\] (cancel out the common terms)
Hence, we get \[(\dfrac{{dy}}{{dx}})[\dfrac{{nx - my}}{{y(x + y)}}] = \dfrac{{nx - my}}{{y(x + y)}} \Rightarrow (\dfrac{{dy}}{{dx}}) = \dfrac{{nx - my}}{{x(x + y)}}[\dfrac{{y(x + y)}}{{nx - my}}]\]
Canceling the common terms, we get \[(\dfrac{{dy}}{{dx}}) = \dfrac{{nx - my}}{{x(x + y)}}[\dfrac{{y(x + y)}}{{nx - my}}] \Rightarrow (\dfrac{{dy}}{{dx}}) = \dfrac{y}{x}\]
Thus, we have \[(\dfrac{{dy}}{{dx}}) = \dfrac{y}{x}\], since from the given question we have $x = 1,y = 2$ and substituting the values we get ${(\dfrac{{dy}}{{dx}})_{x = 1,y = 2}} = \dfrac{y}{x} = \dfrac{2}{1} \Rightarrow 2$
Therefore, the option $B)2$ is correct.

Note: Differentiation and integration are inverse processes like a derivative of \[\dfrac{{d({x^2})}}{{dx}} = 2x\] and the integration is $\int {2xdx = \dfrac{{2{x^2}}}{2}} \Rightarrow {x^2}$
The main concept used in the given problem is the chain rule and derivative of the logarithm function so we used $\log {x^m} = m\log x$ and logarithm derivative function can be represented as $\dfrac{d}{{dx}}\log x = \dfrac{1}{x}$
And chain rule can be represented as $\dfrac{d}{{dx}}(f(g(x)) = {f^1}(g(x)) \times {g^1}(x)$ where \[f(x) = (m + n),g(x) = \log (x + y)\]