How do you find $\dfrac{{dy}}{{dx}}$ given $y + {y^3} = {x^2}$?
Answer
559.2k+ views
Hint: In the given question, we are required to find the derivative of the function provided to us. So, we will differentiate both sides of the equation given to us with respect to x using the power rule of differentiation. Then, we will take the $\dfrac{{dy}}{{dx}}$ term common from the left side of the equation and find its value using the method of transposition.
Complete step by step answer:
We are given the equation $y + {y^3} = {x^2}$ and we have to find the value of $\dfrac{{dy}}{{dx}}$.
So, we differentiate both sides of the equation with respect to x.
$ \Rightarrow \dfrac{d}{{dx}}\left( {y + {y^3}} \right) = \dfrac{d}{{dx}}\left( {{x^2}} \right)$
Now, we know that $\dfrac{d}{{dx}}\left( {a + b} \right) = \dfrac{d}{{dx}}\left( a \right) + \dfrac{d}{{dx}}\left( b \right)$.
So, we get,
$ \Rightarrow \dfrac{d}{{dx}}\left( y \right) + \dfrac{d}{{dx}}\left( {{y^3}} \right) = \dfrac{d}{{dx}}\left( {{x^2}} \right)$
Using the power rule of differentiation $\dfrac{{d\left( {{x^n}} \right)}}{{dx}} = n{x^{n - 1}}$, we get,
$ \Rightarrow \dfrac{d}{{dx}}\left( y \right) + \dfrac{d}{{dx}}\left( {{y^3}} \right) = 2\left( {{x^{2 - 1}}} \right)$
Now, we will follow the chain rule of differentiation $\dfrac{{d\left( {fog\left( x \right)} \right)}}{{dx}} = f'\left( {g\left( x \right)} \right)g'\left( x \right)$.
$ \Rightarrow \dfrac{{dy}}{{dx}} + \left( {3{y^{3 - 1}}} \right)\dfrac{{dy}}{{dx}} = 2\left( {{x^{2 - 1}}} \right)$
Simplifying the expression,
$ \Rightarrow \dfrac{{dy}}{{dx}} + 3{y^2}\dfrac{{dy}}{{dx}} = 2x$
Taking $\dfrac{{dy}}{{dx}}$ common from the left side of the equation, we get,
$ \Rightarrow \dfrac{{dy}}{{dx}}\left( {1 + 3{y^2}} \right) = 2x$
Dividing both the sides of the equation by $\left( {1 + 3{y^2}} \right)$, we get,
$ \therefore \dfrac{{dy}}{{dx}} = \dfrac{{2x}}{{\left( {1 + 3{y^2}} \right)}}$
So, the value of $\dfrac{{dy}}{{dx}}$ is $\dfrac{{2x}}{{\left( {1 + 3{y^2}} \right)}}$.
Note: Chain rule of differentiation $\dfrac{{d\left( {fog\left( x \right)} \right)}}{{dx}} = f'\left( {g\left( x \right)} \right)g'\left( x \right)$ helps us to differentiate the composite and complex functions layer by layer. Power rule of differentiation helps to differentiate the power function as ${x^n}$ with respect to x. Transposition rule states that both sides of the equation remain equal if we multiply, divide, add or subtract the same quantity on both sides.
Complete step by step answer:
We are given the equation $y + {y^3} = {x^2}$ and we have to find the value of $\dfrac{{dy}}{{dx}}$.
So, we differentiate both sides of the equation with respect to x.
$ \Rightarrow \dfrac{d}{{dx}}\left( {y + {y^3}} \right) = \dfrac{d}{{dx}}\left( {{x^2}} \right)$
Now, we know that $\dfrac{d}{{dx}}\left( {a + b} \right) = \dfrac{d}{{dx}}\left( a \right) + \dfrac{d}{{dx}}\left( b \right)$.
So, we get,
$ \Rightarrow \dfrac{d}{{dx}}\left( y \right) + \dfrac{d}{{dx}}\left( {{y^3}} \right) = \dfrac{d}{{dx}}\left( {{x^2}} \right)$
Using the power rule of differentiation $\dfrac{{d\left( {{x^n}} \right)}}{{dx}} = n{x^{n - 1}}$, we get,
$ \Rightarrow \dfrac{d}{{dx}}\left( y \right) + \dfrac{d}{{dx}}\left( {{y^3}} \right) = 2\left( {{x^{2 - 1}}} \right)$
Now, we will follow the chain rule of differentiation $\dfrac{{d\left( {fog\left( x \right)} \right)}}{{dx}} = f'\left( {g\left( x \right)} \right)g'\left( x \right)$.
$ \Rightarrow \dfrac{{dy}}{{dx}} + \left( {3{y^{3 - 1}}} \right)\dfrac{{dy}}{{dx}} = 2\left( {{x^{2 - 1}}} \right)$
Simplifying the expression,
$ \Rightarrow \dfrac{{dy}}{{dx}} + 3{y^2}\dfrac{{dy}}{{dx}} = 2x$
Taking $\dfrac{{dy}}{{dx}}$ common from the left side of the equation, we get,
$ \Rightarrow \dfrac{{dy}}{{dx}}\left( {1 + 3{y^2}} \right) = 2x$
Dividing both the sides of the equation by $\left( {1 + 3{y^2}} \right)$, we get,
$ \therefore \dfrac{{dy}}{{dx}} = \dfrac{{2x}}{{\left( {1 + 3{y^2}} \right)}}$
So, the value of $\dfrac{{dy}}{{dx}}$ is $\dfrac{{2x}}{{\left( {1 + 3{y^2}} \right)}}$.
Note: Chain rule of differentiation $\dfrac{{d\left( {fog\left( x \right)} \right)}}{{dx}} = f'\left( {g\left( x \right)} \right)g'\left( x \right)$ helps us to differentiate the composite and complex functions layer by layer. Power rule of differentiation helps to differentiate the power function as ${x^n}$ with respect to x. Transposition rule states that both sides of the equation remain equal if we multiply, divide, add or subtract the same quantity on both sides.
Recently Updated Pages
Which of the following graphs shows the variation of class 12 physics CBSE

Draw a labelled diagram of the human male reproductive class 12 biology CBSE

Describe the experiment to compare the emf of two cells class 12 physics CBSE

What is standard hydrogen electrode

What is conventional current and electric current class 12 physics CBSE

2Bromopentane is treated with an alcoholic KOH solution class 12 chemistry CBSE

Trending doubts
Draw a labelled sketch of the human eye class 12 physics CBSE

Which are the Top 10 Largest Countries of the World?

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Draw ray diagrams each showing i myopic eye and ii class 12 physics CBSE

Which is the correct genotypic ratio of mendel dihybrid class 12 biology CBSE

What is the Full Form of PVC, PET, HDPE, LDPE, PP and PS ?

