Implicit Differentiation of 5=2x^2y+7xy^2

[karush]

find dy/dx of
$$5=2x^2y+7xy^2$$

ok didn't know how to set this up with the $5=$

 

[topsquark]

find dy/dx of
$$5=2x^2y+7xy^2$$

ok didn't know how to set this up with the $5=$

When you take the derivative of both sides you wind up with \(\displaystyle \dfrac{d}{dx}(5)\). What is the derivative of a constant?

-Dan

 

[karush]

5x

But how do we know it is x not y

 

[topsquark]

5x

But how do we know it is x not y

You are trying to find the derivative \(\displaystyle \dfrac{dy}{dx}\) right?

\(\displaystyle 5 = 2x^2 y + 7x y^2\)

\(\displaystyle \dfrac{d}{dx}(5) = \dfrac{d}{dx} (2x^2 y) + \dfrac{d}{dx} (7x y^2) \)

-Dan

 

[HOI]

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]

$\displaystyle \dfrac{d}{dx}(5) = \dfrac{d}{dx} (2x^2 y) + \dfrac{d}{dx} (7x y^2)$
d/dx
$0=4xy-2x^2yy'+7y^2-14xyy'$
isolate y'
$-2x^2yy'-14xyy'=4xy+7y^2$
$y'(-2x^2y-14xy)=4xy+7y^2$
simplify
$\displaystyle y'= \frac{(4 xy + 7 y^2)}{(-2x^2 - 14x y)}=- \frac{y (4 x + 7 y)}{2 x (x + 7 y)}$

 

[skeeter]

$\displaystyle \dfrac{d}{dx}(5) = \dfrac{d}{dx} (2x^2 y) + \dfrac{d}{dx} (7x y^2)$
d/dx
$0=4xy {\color{red}{+} 2x^2 y'}+7y^2 {\color{red}{+}} 14xyy'$

corrections ... try again

 

[karush]

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
$\displaystyle
y'= \frac{(4 xy + 7 y^2)}{(-2x^2 - 14x y)}
=- \frac{y (4 x + 7 y)}{2 x (x + 7 y)}$

 

[topsquark]

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
$\displaystyle
y'= \frac{(4 xy + 7 y^2)}{(-2x^2 - 14x y)}
=- \frac{y (4 x + 7 y)}{2 x (x + 7 y)}$

Look more carefully at skeeter's post.
\(\displaystyle \dfrac{d}{dx}(2x^2 y) = 4xy + 2x^2y'\)

\(\displaystyle \dfrac{d}{dx}(7x y^2) = 7y^2 + 14xy y'\)

Why are you putting the negative sign in there?

-Dan

 

[karush]

like this
$0=4xy+2x^2y'+7y^2+14xyy'$
isolate y'
$-2x^2y'-14xyy'=4xy+7y^2$
$y'(-2x^2-14xy)=4xy+7y^2$
simplify
$\displaystyle
y'= \frac{(4 xy + 7 y^2)}{(-2x^2 - 14x y)}
=- \frac{y (4 x + 7 y)}{2 x (x + 7 y)}$