Implicit Differentiation

For f(x,y):

$\dfrac {df} {dx} = \dfrac {df}{dx}\dfrac {dx}{dx} + \dfrac {df}{dy} \dfrac {fy}{dx}$

As you might guess, $\dfrac {dx}{dx} = 1$ .

$f(x,y) = g(x,y)$

$\dfrac {d}{dx} f(x,y) = \dfrac {d}{dx} g(x,y)$

We will do the big operation shown above to each side of the equality.

It may help to show it on example pieces:

f(x,y) = y

$\dfrac {d(y)}{dy} \dfrac {dy}{dx} = 1 \dfrac {dy}{dx}$

If we can move things around such that we end up with a single y on one side, implicit differentiation gives us dy/dx on that side.

Now, assume that g(x,y) only has constants and x variables in it, making it g(x).

$\dfrac {d(g)}{dx}\dfrac {dx}{dx} + \dfrac {d(g)}{dy} \dfrac {fy}{dx} = \dfrac {d(g)}{dx} 1 + 0 \dfrac {dy}{dx}$

That variable ‘g’ could have been any variable, and it might be helpful to us to change it to ‘f’. We can then write the following:

$y = f(x)$

$\dfrac {dy}{dx} = \dfrac {d}{dx} f(x)$

The above work shows us the advantage of getting ‘y’ by itself on one side and everything on the other side being all ‘x’ variables and constants.

Appendix A

On a test you will probably get a problem that has x’s and y’s together. Instructions might just simply say you have to solve this using implicit differentiation.

$x^2 + y^2 = 25$

$\dfrac {d (x^2 + y^2)} {dx} \dfrac {dx} {dx} + \dfrac {d (x^2 + y^2)} {dy} \dfrac {dy} {dx}$
$2x (1) + 2y \dfrac {dy}{dx}$

Any derivative of a constant is going to be zero.

$2x + 2y \dfrac {dy}{dx} = 0$

$2y \dfrac {dy}{dx} = -2x$

$\dfrac {dy}{dx} =- \dfrac {2x}{2y}$

$\dfrac {dy}{dx} =- \dfrac {x}{y}$

The above answer is correct as is, and it may provide insight to look at it the way it is. We will also follow the convention of trying to have everything on the right side exprusing only ‘x’ variables, so we rewrite it as the following:

$\dfrac {dy}{dx} =- \dfrac {x}{ \sqrt{25 - x^2}}$

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