Question

If \[{\log _{10}}\left( {\dfrac{{{x^3} - {y^3}}}{{{x^3} + {y^3}}}} \right) = 2\] , then show that \[\dfrac{{dy}}{{dx}} = - \dfrac{{99{x^2}}}{{101{y^2}}}\] .

Answer 1

Hint: We solve this by differentiation. We also know the property of logarithm that is if we have \[{\log _{10}}y = x\] , taking antilog on both side we get \[y = anti{\text{ }}{\log _{10}}(x)\] . We also know the differentiation formula for the power rule, \[\dfrac{d}{{dx}}({x^n}) = n{x^{n - 1}}\] that is with respect to ‘x’. Using these we can prove the given problem.

Complete step-by-step answer:
Given, \[{\log _{10}}\left( {\dfrac{{{x^3} - {y^3}}}{{{x^3} + {y^3}}}} \right) = 2\]
Now taking anti logarithmic on the both sides of the above equation we get,
 \[ \Rightarrow \dfrac{{{x^3} - {y^3}}}{{{x^3} + {y^3}}} = anti{\text{ }}{\log _{10}}(2)\]
Also we know the value of, \[anti{\text{ }}{\log _{10}}(2) = {10^2} = 100\] . Then above becomes:
 \[ \Rightarrow \dfrac{{{x^3} - {y^3}}}{{{x^3} + {y^3}}} = 100\]
Multiplying \[{x^3} + {y^3}\] on both sides, we get:
 \[ \Rightarrow {x^3} - {y^3} = 100 \times ({x^3} + {y^3})\]
Expanding the brackets, we have:
 \[ \Rightarrow {x^3} - {y^3} = 100{x^3} + 100{y^3}\]
Separating ‘y’ terms and ‘x’ terms we have,
 \[ \Rightarrow - 100{y^3} - {y^3} = 100{x^3} - {x^3}\]
Taking \[{y^3}\] common on the left hand side and \[{x^3}\] common on the right hand side, we have:
 \[ \Rightarrow - {y^3}(100 + 1) = (100 - 1){x^3}\]
Adding and subtracting terms in brackets,
 \[ \Rightarrow - 101{y^3} = 99{x^3}\]
Now differentiating with respect to ‘x’ we have:
 \[ \Rightarrow \dfrac{d}{{dx}}\left( { - 101{y^3}} \right) = \dfrac{d}{{dx}}\left( {99{x^3}} \right)\]
Using constant multiple rule of differentiation, we take constant term outside, we have:
 \[ \Rightarrow - 101\dfrac{d}{{dx}}\left( {{y^3}} \right) = 99\dfrac{d}{{dx}}\left( {{x^3}} \right)\]
Differentiating we have,
 \[ \Rightarrow - 101\left( {3{y^2}\dfrac{{dy}}{{dx}}} \right) = 99\left( {3{x^2}\dfrac{{dx}}{{dx}}} \right)\]
Cancelling 3 on both sides and also dx will cancels out,
 \[ \Rightarrow - 101{y^2}\dfrac{{dy}}{{dx}} = 99{x^2}\]
Divided by \[ - 101{y^2}\] on both sides.
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{{99{x^2}}}{{101{y^2}}}\] .
Hence, the solution.
So, the correct answer is “Option C”.

Note: In above we differentiate the equation involving ‘x’ and ‘y’. Which is called as implicit differentiation. Using this we can differentiate the equation of circle, equation of parabola, equation of hyperbola etc.. Know all the properties of logarithms. We can find the antilogarithm value using the common Antilogarithm table. We know that differentiation of constant is zero, meaning that it will be a horizontal line with slope zero. Careful in the differentiation of the function with respect to ‘x’ or ‘y’.