Question

How to find \[\dfrac{{dy}}{{dx}}\] if \[y = \ln \left( {8{x^2} + 9{y^2}} \right)\]

Answer 1

Hint:
Here, we have to find the derivative of the given function. We will use the derivative formula to find the derivative of the logarithmic function. Then we will find the derivative of the algebraic function by using the concept of Implicit differentiation. We will simplify the equation further to get the required answer.

Formula Used:
We will use the following formulas:
1) Derivative formula: \[\dfrac{d}{{dx}}\left( {\ln x} \right) = \dfrac{1}{x}\]
2) Derivative formula: \[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\]

Complete step by step solution:
We are given with a function \[y = \ln \left( {8{x^2} + 9{y^2}} \right)\]
Now, we will find the derivative of the given function.
Now, we will find the derivative of the logarithmic function followed by the derivative of the algebraic function simultaneously.
Using the derivative formula \[\dfrac{d}{{dx}}\left( {\ln x} \right) = \dfrac{1}{x}\], we get
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{8{x^2} + 9{y^2}}}\left[ {\dfrac{d}{{dx}}\left( {8{x^2} + 9{y^2}} \right)} \right]\]
Simplifying the equation, we get
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{8{x^2} + 9{y^2}}}\dfrac{d}{{dx}}\left( {8{x^2}} \right) + \dfrac{1}{{8{x^2} + 9{y^2}}}\dfrac{d}{{dx}}\left( {9{y^2}} \right)\]
Now, by using the derivative formula \[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\], we get
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{8 \cdot 2x}}{{8{x^2} + 9{y^2}}} + \dfrac{{9 \cdot 2y}}{{8{x^2} + 9{y^2}}}\dfrac{{dy}}{{dx}}\]
Multiplying the terms, we get
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{16x}}{{8{x^2} + 9{y^2}}} + \dfrac{{18y}}{{8{x^2} + 9{y^2}}}\dfrac{{dy}}{{dx}}\]
Rewriting the equation, we get
\[ \Rightarrow \dfrac{{dy}}{{dx}} - \dfrac{{18y}}{{8{x^2} + 9{y^2}}}\dfrac{{dy}}{{dx}} = \dfrac{{16x}}{{8{x^2} + 9{y^2}}}\]
Now, by taking out the common factor, we get
\[ \Rightarrow \dfrac{{dy}}{{dx}}\left( {1 - \dfrac{{18y}}{{8{x^2} + 9{y^2}}}} \right) = \dfrac{{16x}}{{8{x^2} + 9{y^2}}}\]
Taking LCM of the terms inside the bracket on the RHS, we get
\[ \Rightarrow \dfrac{{dy}}{{dx}}\left( {1 \times \dfrac{{8{x^2} + 9{y^2}}}{{8{x^2} + 9{y^2}}} - \dfrac{{18y}}{{8{x^2} + 9{y^2}}}} \right) = \dfrac{{16x}}{{8{x^2} + 9{y^2}}}\]
\[ \Rightarrow \dfrac{{dy}}{{dx}}\left( {\dfrac{{8{x^2} + 9{y^2} - 18y}}{{8{x^2} + 9{y^2}}}} \right) = \dfrac{{16x}}{{8{x^2} + 9{y^2}}}\]
Now, by rewriting the terms, we get
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\dfrac{{16x}}{{8{x^2} + 9{y^2}}}}}{{\left( {\dfrac{{8{x^2} + 9{y^2} - 18y}}{{8{x^2} + 9{y^2}}}} \right)}}\]
Cancelling out the same terms of the fractions, we get
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{16x}}{{8{x^2} + 9{y^2} - 18y}}\]

Therefore, the derivative \[\dfrac{{dy}}{{dx}}\] of the function \[y = \ln \left( {8{x^2} + 9{y^2}} \right)\] is \[\dfrac{{16x}}{{8{x^2} + 9{y^2} - 18y}}\].

Note:
We know that Differentiation is a method of finding the derivative of a function and finding the rate of change of function with respect to one variable. But here, we are using the concept of Implicit differentiation. Implicit Differentiation is a process of finding the derivative of a function when the function has both the terms\[x\] and\[y\]. Implicit Differentiation is similar to the process of differentiation and uses the same formula used for differentiation.