# Thread: evaluate y' at (2,1)

1. ## evaluate y' at (2,1)

For the equation given below, evaluate at the point .

y' at (2,1) = ?

so i tried taking the natural log of everything then differentiating both sides and solving for y'.

ln2 + lne^xy - ln2 + lnx = lny +ln9.78

xy + lnx = lny + ln9.78

(1/y)y' = y + xy' + 1/x + 0

y' = y^2 + xy'y + y/x

y' = y^2+(y/x) / 1-xy

then if you plug in (2,1) i get -1.5 and thats not right =(

2. Originally Posted by mybrohshi5
For the equation given below, evaluate at the point .

y' at (2,1) = ?

so i tried taking the natural log of everything then differentiating both sides and solving for y'.

ln2 + lne^xy - ln2 + lnx = lny +ln9.78

any advice? yes, the entire step above is invalid

ln(a+b) does not equal ln(a) + ln(b)
...

3. Also, you can't take the log of everything and then differentiate as you would be differentiating $log(y)$ and not $y$ as required..

4. ok i guess i have not a clue what to do then.

should i start off by taking the natural log of both sides?

then differentiating with u'/u for both sides?

then solving for y'?

thank you

5. why log? just take the derivative straight up.

$\frac{d}{dx}(2e^{xy} - 2x = y + k)$

$2e^{xy}(xy' + y) - 2 = y'$

$2xy'e^{xy} + 2ye^{xy} - 2 = y'$

$2ye^{xy} - 2 = y' - 2xy'e^{xy}$

$2ye^{xy} - 2 = y'(1 - 2xe^{xy})$

$\frac{2(ye^{xy}-1)}{1 - 2xe^{xy}} = y'$

6. Originally Posted by mybrohshi5
ok i guess i have not a clue what to do then.

should i start off by taking the natural log of both sides?

then differentiating with u'/u for both sides?

then solving for y'?

thank you
You shouldn't have to take logarithms. You can just differentiate implicity.

$2e^{xy}-2x=y+9.78$

The derivative of $2e^{xy}$ is kind of tricky. Set $u=xy$ and differentiate as follows:

$2e^{u}\frac{du}{dx}$

$\frac{du}{dx}=x\frac{dy}{dx}+y$ (by product rule)

$=2e^u(\frac{xdy}{dx}+y)$

$=2e^{xy}(\frac{xdy}{dx}+y)$

So now that you know the derivative of this term, the rest should be obvious.