Implicit Differentiation Help

[skyblue]

The slope of the curve y^2 - xy = 3 at the point where y=1 is....


so far i did 2y(dy/dx) - x- y(dy/dx) = 0

dy/dx(2y-y) = -x
(2y-y)/(2y-y) = -x/(2y-y)

now i don't understand the y=1

 

[pka]

\(\displaystyle \L\begin{array}{l}
y^2 - xy = 3 \\
2yy' - y - xy' = 0 \\
y' = \frac{y}{{2y - x}} \\
\end{array}\)

 

[skyblue]

hi thank you. but how would I find the slope of the curve where y=1.

 

[pka]

When y=1 then x=? and y'=?

 

[skeeter]

substitute y = 1 into the original equation and solve for x ... then use that value for x and 1 for y and substitute both into the derivative expression to find the slope.

 

[skyblue]

ohh i see. so it would be -2. thanks for all of your help.

 

[skyblue]

can someone also please help me with:

radical x + radical y = 4y^2 , then dy/dx at (1,1) =

this is what i tried so far: (1/2)x + (1/2)y(dy/dx) = 8y(dy/dx)
and i'm not sure how to evaluate it at (1,1)

 

[pka]

\(\displaystyle \L\sqrt x + \sqrt y = 4y^2 \quad \Rightarrow \quad \frac{1}{{2\sqrt x }} + \frac{{y'}}{{2\sqrt y }} = 8yy'\)
Solve for \(\displaystyle y'\) and substitute.

 

[skeeter]

it also looks like you do not know how to find derivatives ...

\(\displaystyle \L \sqrt{x} + \sqrt{y} = 4y^2\)

\(\displaystyle \L \frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \frac{dy}{dx} = 8y \frac{dy}{dx}\)

substitute the given values for x and y and solve for \(\displaystyle \L \frac{dy}{dx}\).
you should get the value of the slope at (1,1) to be 1/15.

btw ... in future, start a new problem with a new thread.

 

[skyblue]

thank you so much for your help!