The discussion focuses on finding the derivative \(\frac{dy}{dx}\) of the equation \(5 = 2x^2y + 7xy^2\) using implicit differentiation. Participants clarify that the derivative of a constant, such as 5, is zero, leading to the equation \(0 = 4xy + 2x^2y' + 7y^2 + 14xyy'\). After isolating \(y'\), the final expression for the derivative is \(\displaystyle y' = -\frac{y(4x + 7y)}{2x(x + 7y)}\). This method demonstrates the application of the product rule and implicit differentiation techniques.
PREREQUISITESStudents studying calculus, mathematics educators, and anyone looking to enhance their understanding of implicit differentiation and its applications in solving equations.
When you take the derivative of both sides you wind up with [math]\dfrac{d}{dx}(5)[/math]. What is the derivative of a constant?karush said:find dy/dx of
$$5=2x^2y+7xy^2$$
ok didn't know how to set this up with the $5=$
You are trying to find the derivative [math]\dfrac{dy}{dx}[/math] right?karush said:5x
But how do we know it is x not y
NO! You appear to be integrating rather than differentiating. The derivative of any constant is 0.karush said:5x
But how do we know it is x not y
karush said:$\displaystyle \dfrac{d}{dx}(5) = \dfrac{d}{dx} (2x^2 y) + \dfrac{d}{dx} (7x y^2)$
d/dx
$0=4xy {\color{red}{+} 2x^2 y'}+7y^2 {\color{red}{+}} 14xyy'$
Look more carefully at skeeter's post.karush said:done try more!
$0=4xy-2x^2y'+7y^2-14xyy'$
isolate y'
$-2x^2y'-14xyy'=4xy+7y^2$
$y'(-2x^2-14xy)=4xy+7y^2$
simplify
$\displaystyle
y'= \frac{(4 xy + 7 y^2)}{(-2x^2 - 14x y)}
=- \frac{y (4 x + 7 y)}{2 x (x + 7 y)}$