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Question: How do you find \(\dfrac{{dy}}{{dx}}\) given \(y + {y^3} = {x^2}\)?...

How do you find dydx\dfrac{{dy}}{{dx}} given y+y3=x2y + {y^3} = {x^2}?

Explanation

Solution

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 dydx\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+y3=x2y + {y^3} = {x^2} and we have to find the value of dydx\dfrac{{dy}}{{dx}}.
So, we differentiate both sides of the equation with respect to x.
⇒ddx(y+y3)=ddx(x2)\Rightarrow \dfrac{d}{{dx}}\left( {y + {y^3}} \right) = \dfrac{d}{{dx}}\left( {{x^2}} \right)
Now, we know that ddx(a+b)=ddx(a)+ddx(b)\dfrac{d}{{dx}}\left( {a + b} \right) = \dfrac{d}{{dx}}\left( a \right) + \dfrac{d}{{dx}}\left( b \right).
So, we get,
⇒ddx(y)+ddx(y3)=ddx(x2)\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 d(xn)dx=nxn−1\dfrac{{d\left( {{x^n}} \right)}}{{dx}} = n{x^{n - 1}}, we get,
⇒ddx(y)+ddx(y3)=2(x2−1)\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 d(fog(x))dx=f′(g(x))g′(x)\dfrac{{d\left( {fog\left( x \right)} \right)}}{{dx}} = f'\left( {g\left( x \right)} \right)g'\left( x \right).
⇒dydx+(3y3−1)dydx=2(x2−1)\Rightarrow \dfrac{{dy}}{{dx}} + \left( {3{y^{3 - 1}}} \right)\dfrac{{dy}}{{dx}} = 2\left( {{x^{2 - 1}}} \right)
Simplifying the expression,
⇒dydx+3y2dydx=2x\Rightarrow \dfrac{{dy}}{{dx}} + 3{y^2}\dfrac{{dy}}{{dx}} = 2x
Taking dydx\dfrac{{dy}}{{dx}} common from the left side of the equation, we get,
⇒dydx(1+3y2)=2x\Rightarrow \dfrac{{dy}}{{dx}}\left( {1 + 3{y^2}} \right) = 2x
Dividing both the sides of the equation by (1+3y2)\left( {1 + 3{y^2}} \right), we get,
∴dydx=2x(1+3y2)\therefore \dfrac{{dy}}{{dx}} = \dfrac{{2x}}{{\left( {1 + 3{y^2}} \right)}}

So, the value of dydx\dfrac{{dy}}{{dx}} is 2x(1+3y2)\dfrac{{2x}}{{\left( {1 + 3{y^2}} \right)}}.

Note: Chain rule of differentiation d(fog(x))dx=f′(g(x))g′(x)\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 xn{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.