Question

If \[{x^{\dfrac{2}{3}}} + {y^{\dfrac{2}{3}}} = {a^{\dfrac{2}{3}}}\] then \[\dfrac{{dy}}{{dx}} = \]
A. \[{\left( {\dfrac{y}{3}} \right)^{\dfrac{1}{3}}}\]
B. \[{\left( { - \dfrac{y}{x}} \right)^{\dfrac{1}{3}}}\]
C. \[{\left( {\dfrac{x}{y}} \right)^{\dfrac{1}{3}}}\]
D. \[{\left( { - \dfrac{x}{y}} \right)^{\dfrac{1}{3}}}\]

Answer 1

Hint: In this question, we try to form the indices formula for the value \[\dfrac{2}{3}\] and take the indices of the form of \[{x^{\dfrac{2}{3}}},{y^{\dfrac{2}{3}}},{a^{\dfrac{2}{3}}}\] then we take the individual differentiation of the function then substitute it into the original function and simplified it to get the desired result.

Formula used:
We have been using the following formula to find the derivative:
1. \[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\]

Complete step-by-step solution:
We are given that \[{x^{\dfrac{2}{3}}} + {y^{\dfrac{2}{3}}} = {a^{\dfrac{2}{3}}}...\left( 1 \right)\]
We know the given function is in cube root forms and the given values are in the form of indices.
Now we find the individual differentiation of the given function
So,
\[\dfrac{d}{{dx}}\left( {{x^{\dfrac{2}{3}}}} \right) + \dfrac{d}{{dx}}\left( {{y^{\dfrac{2}{3}}}} \right) = \dfrac{d}{{dx}}\left( {{a^{\dfrac{2}{3}}}} \right)...\left( 2 \right)\]
Now we use the derivative formula \[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\], we have
First, we take derivative of \[\dfrac{d}{{dx}}\left( {{x^{\dfrac{2}{3}}}} \right)\], we get
\[
  \dfrac{d}{{dx}}\left( {{x^{\dfrac{2}{3}}}} \right) = \dfrac{2}{3}{x^{\dfrac{2}{3} - 1}} \\
   = \dfrac{2}{3}\,{x^{\dfrac{{2 - 3}}{3}}} \\
   = \dfrac{2}{3}{x^{ - \dfrac{1}{3}}}...\left( 3 \right) \\
 \]
Now we take the derivative of \[\dfrac{d}{{dx}}\left( {{y^{\dfrac{2}{3}}}} \right)\], we get
\[
  \dfrac{d}{{dx}}\left( {{y^{\dfrac{2}{3}}}} \right) = \dfrac{2}{3}{y^{\dfrac{2}{3} - 1}}\dfrac{{dy}}{{dx}} \\
   = \dfrac{2}{3}\,{y^{\dfrac{{2 - 3}}{3}}}\dfrac{{dy}}{{dx}} \\
   = \dfrac{2}{3}\,{y^{ - \dfrac{1}{3}}}\dfrac{{dy}}{{dx}}...\left( 4 \right) \\
 \]
Now we take the derivative of \[\dfrac{d}{{dx}}\left( {{a^{\dfrac{2}{3}}}} \right)\], we get
\[
  \dfrac{d}{{dx}}\left( {{a^{\dfrac{2}{3}}}} \right) = 0..\left( 5 \right)
 \] (because a is constant)
Now we substitute equation (3), (4), (5) in equation (1), we get
\[
  \dfrac{2}{3}{x^{ - \dfrac{1}{3}}} + \dfrac{2}{3}{y^{ - \dfrac{1}{3}}}\dfrac{{dy}}{{dx}} = 0 \\
  \dfrac{2}{3}\left[ {{x^{ - \dfrac{1}{3}}} + {y^{ - \dfrac{1}{3}}}\dfrac{{dy}}{{dx}}} \right] = 0 \\
  {x^{ - \dfrac{1}{3}}} + {y^{ - \dfrac{1}{3}}}\dfrac{{dy}}{{dx}} = 0 \\
  {y^{ - \dfrac{1}{3}}}\dfrac{{dy}}{{dx}} = - {x^{ - \dfrac{1}{3}}} \\
 \]
Further simplifying, we get
  \[
  \dfrac{{dy}}{{dx}} = \dfrac{{ - {x^{ - \dfrac{1}{3}}}}}{{{y^{ - \dfrac{1}{3}}}}} \\
   = \dfrac{{ - \dfrac{1}{{{x^{\dfrac{1}{3}}}}}}}{{\dfrac{1}{{{y^{\dfrac{1}{3}}}}}}} \\
   = - \dfrac{1}{{{x^{\dfrac{1}{3}}}}} \times \dfrac{{{y^{\dfrac{1}{3}}}}}{1} \\
   = - \dfrac{{{y^{\dfrac{1}{3}}}}}{{{x^{\dfrac{1}{3}}}}} \\
   = {\left[ { - \dfrac{y}{x}} \right]^{\dfrac{1}{3}}}
 \]
Therefore, if \[{x^{\dfrac{2}{3}}} + {y^{\dfrac{2}{3}}} = {a^{\dfrac{2}{3}}}\] then \[\dfrac{{dy}}{{dx}}\]is \[{\left[ {\dfrac{{ - y}}{x}} \right]^{\dfrac{1}{3}}}\].
Hence, option (B) is the correct answer

Note: Students must be careful when determining the derivative \[\dfrac{{dy}}{{dx}}\] because there is a chance that you will make a sign mistake. Also, be cautious when checking the options and substituting the values of \[\dfrac{{dy}}{{dx}}\]. The best way to answer this type of question is to go through each option one by one because it contains \[\dfrac{{dy}}{{dx}}\] terms that cannot be found otherwise.