# Thread: Implicit differentiation with quotients

1. ## Implicit differentiation with quotients

Find $\displaystyle \frac {dy} {dx}$ by implicit differentiation.

$\displaystyle \tan \frac {x} {y}=x+y$

By the chain rule, the left side becomes

$\displaystyle \sec^2 (\frac {x} {y}) *$ (the derivative of $\displaystyle \frac {x} {y}$)

I have two problems, one, where do the $\displaystyle y$-primes go, and two, how do I solve for it?

The final answer should look like

$\displaystyle y'= \frac {y \sec^2 \frac {x} {y} - y^2} {y^2 + x \sec^2 \frac {x} {y}}$

2. Originally Posted by Chokfull
Find $\displaystyle \frac {dy} {dx}$ by implicit differentiation.

$\displaystyle \tan \frac {x} {y}=x+y$

By the chain rule, the left side becomes

$\displaystyle \sec^2 (\frac {x} {y}) *$ (the derivative of $\displaystyle \frac {x} {y}$)

I have two problems, one, where do the $\displaystyle y$-primes go, and two, how do I solve for it?

The final answer should look like

$\displaystyle y'= \frac {y \sec^2 \frac {x} {y} - y^2} {y^2 + x \sec^2 \frac {x} {y}}$

$\displaystyle \sec^2 (\frac {x} {y}) \frac{y-xy'}{y^2} = 1 + y'$

Now we need to expand and group y' terms on one side. I'll try to move the y' terms on right hand side:

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

Expand on left side:

$\displaystyle y \sec^2 (\frac {x} {y})-x\sec^2 (\frac {x} {y})y' = y^2 + y^2y'$

Move y' terms on the right and y^2 from right to left:

$\displaystyle y \sec^2 (\frac {x} {y}) - y^2 = y^2y' +x\sec^2 (\frac {x} {y})y'$

Factor a y', then divide to solve for it. I hope my algebra does not contain any errors above.

Good luck!

3. Originally Posted by apcalculus
$\displaystyle \sec^2 (\frac {x} {y}) \frac{y-xy'}{y^2} = 1 + y'$

Now we need to expand and group y' terms on one side. I'll try to move the y' terms on right hand side:

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

Expand on left side:

$\displaystyle y \sec^2 (\frac {x} {y})-x\sec^2 (\frac {x} {y})y' = y^2 + y^2y'$

Move y' terms on the right and y^2 from right to left:

$\displaystyle y \sec^2 (\frac {x} {y}) - y^2 = y^2y' +x\sec^2 (\frac {x} {y})y'$

Factor a y', then divide to solve for it. I hope my algebra does not contain any errors above.

Good luck!

OK thanks a lot! It looks like i had the derivative right, then I was mainly going the wrong way in my algebra