When you take the derivative of both sides you wind up with \(\displaystyle \dfrac{d}{dx}(5)\). 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 \(\displaystyle \dfrac{dy}{dx}\) 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
$\displaystyle
y'= \frac{(4 xy + 7 y^2)}{(-2x^2 - 14x y)}
=- \frac{y (4 x + 7 y)}{2 x (x + 7 y)}$
"Implicit dy/dx" refers to the derivative of a function that is expressed implicitly, meaning that the dependent variable is not explicitly written in terms of the independent variable. In other words, the function is not in the form y = f(x).
The main difference between "implicit dy/dx" and "explicit dy/dx" is that in implicit differentiation, the dependent variable is not isolated on one side of the equation, making it more challenging to find the derivative. In explicit differentiation, the dependent variable is explicitly written in terms of the independent variable, making it easier to find the derivative.
Functions that involve radicals, logarithms, and trigonometric functions often require implicit differentiation. For example, y^3 + x^2 = 5, ln(y) = x, and sin(xy) = 3x + 2y.
To find the derivative using implicit differentiation, you first take the derivative of both sides of the equation with respect to the independent variable. Then, you use the chain rule to differentiate any terms that involve the dependent variable. Finally, solve for dy/dx to get the derivative.
Implicit differentiation is useful because it allows us to find the derivative of a function that cannot be easily solved for y in terms of x. It is also essential in solving problems involving related rates, optimization, and implicit equations.