Question

question
if ( y - y^{2}+2 x^{3}=0 ) then find ( \frac{d y}{d x} ) at the point ( (1,2) ).
answer attempt 1 out of 2
( left.\frac{d y}{d x}
ight|_{(1,2)}= )

Explanation:

Step1: Differentiate both sides with respect to \(x\)

Differentiate \(y - y^{2}+2x^{3}=0\) term - by - term.
Using the sum rule \(\frac{d}{dx}(u + v+w)=\frac{du}{dx}+\frac{dv}{dx}+\frac{dw}{dx}\), where \(u = y\), \(v=-y^{2}\), \(w = 2x^{3}\).
The derivative of \(y\) with respect to \(x\) is \(\frac{dy}{dx}\) (by the chain rule, since \(y\) is a function of \(x\)).
The derivative of \(-y^{2}\) with respect to \(x\) is \(-2y\frac{dy}{dx}\) (using the chain rule \(\frac{d}{dx}(f(g(x)))=f^{\prime}(g(x))\cdot g^{\prime}(x)\), where \(f(u)=-u^{2}\), \(u = y\), \(f^{\prime}(u)=-2u\), \(g(x)=y\), \(g^{\prime}(x)=\frac{dy}{dx}\)).
The derivative of \(2x^{3}\) with respect to \(x\) is \(6x^{2}\) (using the power rule \(\frac{d}{dx}(ax^{n})=nax^{n - 1}\), here \(a = 2\), \(n = 3\)).
So, \(\frac{dy}{dx}-2y\frac{dy}{dx}+6x^{2}=0\).

Step2: Solve for \(\frac{dy}{dx}\)

Factor out \(\frac{dy}{dx}\) from the left - hand side: \(\frac{dy}{dx}(1 - 2y)+6x^{2}=0\).
Then \(\frac{dy}{dx}(1 - 2y)=-6x^{2}\), and \(\frac{dy}{dx}=\frac{6x^{2}}{2y - 1}\).

Step3: Substitute \(x = 1\) and \(y = 2\)

Substitute \(x = 1\) and \(y = 2\) into \(\frac{dy}{dx}=\frac{6x^{2}}{2y - 1}\).
We get \(\frac{dy}{dx}=\frac{6\times1^{2}}{2\times2 - 1}=\frac{6}{4 - 1}\).

Answer:

\(2\)