Question

If ${x^3}{y^5} = {\left( {x + y} \right)^8}$, then show that $\dfrac{{dy}}{{dx}} = \dfrac{y}{x}$

Answer 1

Hint: In this question, we will proceed by differentiating on both sides w.r.t ‘\[x\]’ by using the product rule of differentiation. Then simplify further by grouping and cancelling the common terms to prove the given equation.

Complete step-by-step answer:
Let the given equation be ${x^3}{y^5} = {\left( {x + y} \right)^8}............................................\left( 1 \right)$
Differentiating equation (1) w.r.t ‘$x$’ on both sides, we have
\[ \Rightarrow \dfrac{d}{{dx}}\left( {{x^3}{y^5}} \right) = \dfrac{d}{{dx}}{\left( {x + y} \right)^8}\]
By product rule we have \[\dfrac{d}{{dx}}\left[ {f\left( x \right)g\left( y \right)} \right] = g\left( y \right)\dfrac{{df\left( x \right)}}{{dx}} + f\left( x \right)\dfrac{{dg\left( y \right)}}{{dx}}\]. So, using product rule we have
\[ \Rightarrow {y^5}\dfrac{d}{{dx}}\left( {{x^3}} \right) + {x^5}\dfrac{d}{{dx}}\left( {{y^5}} \right) = \dfrac{d}{{dx}}{\left( {x + y} \right)^8}\]
We know that \[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\]. By using this formula, we have
\[
   \Rightarrow {y^5}\left( {3{x^2}} \right) + {x^3}\left( {5{y^4}\dfrac{{dy}}{{dx}}} \right) = 8{\left( {x + y} \right)^7}\left[ {\dfrac{d}{{dx}}\left( {x + y} \right)} \right] \\
   \Rightarrow 3{x^2}{y^5} + 5{x^3}{y^4}\dfrac{{dy}}{{dx}} = 8{\left( {x + y} \right)^7}\left[ {\dfrac{d}{{dx}}\left( x \right) + \dfrac{d}{{dx}}\left( y \right)} \right] \\
   \Rightarrow 3{x^2}{y^5} + 5{x^3}{y^4}\dfrac{{dy}}{{dx}} = 8{\left( {x + y} \right)^7}\left[ {1 + \dfrac{{dy}}{{dx}}} \right] \\
\]
Simplifying further, we have
\[
   \Rightarrow 3{x^2}{y^5} + 5{x^3}{y^4}\dfrac{{dy}}{{dx}} = 8{\left( {x + y} \right)^7} + 8{\left( {x + y} \right)^7}\dfrac{{dy}}{{dx}} \\
   \Rightarrow \left( {5{x^3}{y^4} - 8{{\left( {x + y} \right)}^7}} \right)\dfrac{{dy}}{{dx}} = 8{\left( {x + y} \right)^7} - 3{x^2}{y^5}.................................................\left( 2 \right) \\
\]
From equation (1) we have
\[
   \Rightarrow {x^3}{y^5} = {\left( {x + y} \right)^8} \\
   \Rightarrow {x^3}{y^4} = \dfrac{{{{\left( {x + y} \right)}^8}}}{y}..........................................\left( 3 \right) \\
\]
Also, from equation (1), we have
\[
   \Rightarrow {x^3}{y^5} = {\left( {x + y} \right)^8} \\
   \Rightarrow {x^2}{y^5} = \dfrac{{{{\left( {x + y} \right)}^8}}}{x}..........................................\left( 4 \right) \\
\]
Substituting equation (3) and (4) in equation (2), we have
\[ \Rightarrow \left( {5\dfrac{{{{\left( {x + y} \right)}^8}}}{y} - 8{{\left( {x + y} \right)}^7}} \right)\dfrac{{dy}}{{dx}} = 8{\left( {x + y} \right)^7} - 3\dfrac{{{{\left( {x + y} \right)}^8}}}{x}\]
Grouping and cancelling the common terms, we have
\[
   \Rightarrow {\left( {x + y} \right)^7}\left( {5\dfrac{{\left( {x + y} \right)}}{y} - 8} \right)\dfrac{{dy}}{{dx}} = {\left( {x + y} \right)^7}\left[ {8 - 3\dfrac{{\left( {x + y} \right)}}{x}} \right] \\
   \Rightarrow {\left( {x + y} \right)^7}\left( {\dfrac{{5x + 5y - 8y}}{y}} \right)\dfrac{{dy}}{{dx}} = {\left( {x + y} \right)^7}\left[ {\dfrac{{8x - 3x - 3y}}{x}} \right] \\
   \Rightarrow \left( {\dfrac{{5x - 3y}}{y}} \right)\dfrac{{dy}}{{dx}} = \left[ {\dfrac{{5x - 3y}}{x}} \right] \\
   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{5x - 3y}}{x} \times \dfrac{y}{{5x - 3y}} \\
  \therefore \dfrac{{dy}}{{dx}} = \dfrac{y}{x} \\
\]
Hence proved.

Note: Product rule of differentiation states the if the functions \[f\left( x \right)\] and \[g\left( y \right)\] are differentiable then \[\dfrac{d}{{dx}}\left[ {f\left( x \right)g\left( y \right)} \right] = g\left( y \right)\dfrac{{df\left( x \right)}}{{dx}} + f\left( x \right)\dfrac{{dg\left( y \right)}}{{dx}}\]. In these types of questions, try to solve the differentiation in a simpler way by not expanding the powers of the variables.