# Thread: find dy/dx of the following

1. ## find dy/dx of the following

Hi all,

Should be an easy one.

Find dy/dx of:

x^2 + y^2 = 49

2. Originally Posted by Joel
Hi all,

Should be an easy one.

Find dy/dx of:

x^2 + y^2 = 49
I could do this but that'd defy the point of helping. If you use implicit differentiation and the chain rule you'll be on the road to success.

If you want more help please post what you've done

3. Originally Posted by Joel
Hi all,

Should be an easy one.

Find dy/dx of:

x^2 + y^2 = 49
Dear Joel,

Derivate the left and right sides of the equation, and use the chain rule.

$\displaystyle x^2+y^2=49$

$\displaystyle \frac{d}{dx}\left(x^2+y^2\right)=\frac{d(49)}{dx}$

Hope you can continue from here.

4. Thanks for your help.

x^2 + y^2 = 49

y^2 = 49 - x^2

y = +- root(49 - x^2)

y = +- ( 49 - x^2 ) ^1/2

y' = 1/2 (49 - x^2)^-1/2 . (-2x)

am I anywhere near on the right track ???

5. Originally Posted by Joel
Thanks for your help.

x^2 + y^2 = 49

y^2 = 49 - x^2

y = +- root(49 - x^2)

y = +- ( 49 - x^2 ) ^1/2

y' = 1/2 (49 - x^2)^-1/2 . (-2x)

am I anywhere near on the right track ???
I would stick with implicit differentiation rather than trying to solve the quadratic equation.

$\displaystyle \frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(49)$

The first and last terms should be easy enough, for the middle term you can differentiate y as though it was x and then multiply by $\displaystyle \frac{dy}{dx}$ (which is the chain rule)

For example $\displaystyle \frac{d}{dx} \left(\frac{1}{3}y^3\right) = y^2 \frac{dy}{dx}$