Answer
384.6k+ views
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.
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)