## implicit differentiation

i hate the name. it looks like we are doing something implicit here. technically, that is the case.

we will extend our arm of power to implicit functions.

first a reminder: an explicit function is defined as y=something something x. all other functions are implicit functions.

now, as I was saying, how do we calculate derivatives of implicit functions? i remembered that was an obstacle when i tried to study calculus.

that is because, now we have to think of y, alongside of x, as something operable. we will illustrate that below.

think of the function xy=1. what is the derivative? for the purposes here, we will not move y to the right. instead, we differentiate both sides.

step one: use the product rule on the left side. xy'+y.

step two: so we have xy'+y=0. now we isolate the variable y' as in algebra.

step three: we have solved for the derivative in terms of both x and y. we can solve for y if we wish and substitute it back.

we will extend our arm of power to implicit functions.

first a reminder: an explicit function is defined as y=something something x. all other functions are implicit functions.

now, as I was saying, how do we calculate derivatives of implicit functions? i remembered that was an obstacle when i tried to study calculus.

that is because, now we have to think of y, alongside of x, as something operable. we will illustrate that below.

think of the function xy=1. what is the derivative? for the purposes here, we will not move y to the right. instead, we differentiate both sides.

step one: use the product rule on the left side. xy'+y.

__the derivative on the right side is just zero.__**(note that i have sued y' to denote the derivative of f(x). it will make this whole thing a lot simpler.)**step two: so we have xy'+y=0. now we isolate the variable y' as in algebra.

step three: we have solved for the derivative in terms of both x and y. we can solve for y if we wish and substitute it back.