Implicit Differentiation

Implicit Differentiation is a topic where we insist on showing consistency. We are doing a rework on this page so that one equation will work with all scenarios possible with terms containing powers of x and y.

We are also searching for justification for two things: 1) The definition of implicit differentiation, 2) The product rule for functions of two variables

We are going to show Implicit Differentiation with respect to x and hopefully when we are done you will be able to figure out how to do it for y.

Assuming you’ve already started learning Differentiation, you started with equations having y on the left side and one or more terms in x on the right side.

y=f(x)

This works for a lot of the math that we use in science (at least for a few years) but we’d like you to be able to do differentiation on an equation with two variables, x and y. You can move terms from one side to the other, so it is possible to move every term with y in it to the right side. Having all terms with y on one side will put all dy/dx factors on one side and you want this, as you will soon see:

f(x)=g(x,y)

We are interested in df/dx and dg/dx and we’ll do something to both of them that involves the Chain Rule:

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

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

…

\dfrac{dg}{dx} = \dfrac{dg}{dx} \dfrac{dx}{dx} + \dfrac{dg}{dy} \dfrac{dy}{dx}

\dfrac{dg}{dx} = \dfrac{d}{dx} g(x,y) \dfrac{dx}{dx} + \dfrac{d}{dy} g(x,y) \dfrac{dy}{dx}

Now that you’ve seen the entire equations, we can let ratios equal to one, such as dx/dx, drop out:

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

\dfrac{dg}{dx} = \dfrac{d}{dx} g(x,y) + \dfrac{d}{dy} g(x,y) \dfrac{dy}{dx}

We can make one more term drop out because the derivative with respect to u zero:

Now if f(x)=g(x,y) then

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

The above is equivalent to the right sides of the two equations, so the left sides must also be equal:

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

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

\begin{bmatrix} \dfrac{d}{dy} f(x) - \dfrac{d}{dy} g(x,y)\end{bmatrix} \dfrac{dy}{dx} = \dfrac{d}{dx} g(x,y) - \dfrac{d}{dx} f(x)

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

Comments

The identity below should be used for f(x,y):

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

  • If there is no y then the second term goes to zero
  • If a term is a constant then both terms go to zero

\dfrac{df}{dx} = \dfrac{df}{dy} \dfrac{dy}{dx}

A Change to Our Thinking

\dfrac{d}{dx}y^2=2y \dfrac {dy}{dx}

There could be a problem that if someone sees a single letter without the parentheses and assumes that single letter represents a variable.

  • y could be a variable.
  • y could be a function.

If we assume y to be a function then we have both scenarios covered.

\dfrac {d}{dx} y^2=2y \dfrac{dy}{dx}

If y is a variable holding a constant, then dy/d