So far, we have only differentiated functions written in the explicit form $y = f(x)$. But many relationships in mathematics are not written this way. For example, how would we find the slope of the tangent line to a circle like $x^2 + y^2 = 25$? Today, we learn a powerful new technique, called implicit differentiation, that allows us to find derivatives in these more complex situations.
We start by comparing an explicit function, like the top half of a circle, $y = \sqrt{25 - x^2}$, with the implicit relation for the full circle, $x^2 + y^2 = 25$.
Why this matters: Many important curves (circles, ellipses, and more complex shapes) are not functions because they fail the vertical line test. It's often difficult or impossible to solve for $y$ explicitly in terms of $x$. Implicit differentiation gives us a powerful tool to find the slope of the tangent line at any point on such a curve without having to solve for $y$ first.
Is the relation $y^2 + x = 4$ defined implicitly or explicitly? Can you rewrite it as one or more explicit functions?
The Core Idea: We operate under the assumption that $y$ is a function of $x$ (i.e., $y = f(x)$) and differentiate both sides of the equation with respect to $x$.
The Golden Rule: Any time you differentiate a term involving $y$, you must apply the Chain Rule, which results in a $\frac{dy}{dx}$ factor. For example, $\frac{d}{dx}(y^3) = 3y^2 \cdot \frac{dy}{dx}$.
The 5-Step Process:
Find $\frac{dy}{dx}$ for the circle $x^2 + y^2 = 25$.
Step 1 & 2: Differentiate both sides with respect to $x$, using the Chain Rule on the $y^2$ term.
$$ \frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(25) $$ $$ 2x + 2y \cdot \frac{dy}{dx} = 0 $$Step 3: Move the term without $\frac{dy}{dx}$ to the other side.
$$ 2y \frac{dy}{dx} = -2x $$Step 4 & 5: Solve for $\frac{dy}{dx}$.
$$ \frac{dy}{dx} = \frac{-2x}{2y} = -\frac{x}{y} $$The graph shows that our formula $\frac{dy}{dx} = -\frac{x}{y}$ for the circle correctly gives the slope at a given point. Use the slider to move the point and see the tangent line change.
Find $\frac{dy}{dx}$ for the relation $x^2 + \cos(y) = xy^3$. It is practically impossible to isolate $y$ in this equation, but we can still find its derivative.
First, we differentiate both sides with respect to $x$.
$$ \frac{d}{dx}(x^2) + \frac{d}{dx}(\cos(y)) = \frac{d}{dx}(xy^3) $$The left side is straightforward, remembering the chain rule for the $y$ term:
$$ 2x - \sin(y) \cdot \frac{dy}{dx} = \frac{d}{dx}(xy^3) $$For the right side, we must use the Product Rule. Let $f(x) = x$ and $g(x) = y^3$. Recall the rule is $(fg)' = f'g + fg'$.
Applying the product rule to $xy^3$ gives:
$$ \frac{d}{dx}(xy^3) = (1)(y^3) + (x)(3y^2 \cdot \frac{dy}{dx}) = y^3 + 3xy^2\frac{dy}{dx} $$Now, we set the differentiated left and right sides equal:
$$ 2x - \sin(y) \cdot \frac{dy}{dx} = y^3 + 3xy^2\frac{dy}{dx} $$Next, gather all terms containing $\frac{dy}{dx}$ on one side and all other terms on the opposite side.
$$ 2x - y^3 = 3xy^2 \frac{dy}{dx} + \sin(y) \frac{dy}{dx} $$Finally, factor out $\frac{dy}{dx}$ and solve for it.
$$ 2x - y^3 = \frac{dy}{dx} (3xy^2 + \sin(y)) $$ $$ \frac{dy}{dx} = \frac{2x - y^3}{3xy^2 + \sin(y)} $$Find $\frac{dy}{dx}$ for the relation $x^3 - xy + y^2 = 4$.
The primary application of finding $\frac{dy}{dx}$ is to determine the slope of the tangent line to a curve at a specific point.
Find the equation of the tangent line to the curve $x^3 + y^3 = 6xy$ at the point $(3, 3)$.
First, find $\frac{dy}{dx}$:
$$ 3x^2 + 3y^2 \frac{dy}{dx} = 6y + 6x \frac{dy}{dx} $$ $$ (3y^2 - 6x)\frac{dy}{dx} = 6y - 3x^2 \implies \frac{dy}{dx} = \frac{6y - 3x^2}{3y^2 - 6x} = \frac{2y-x^2}{y^2-2x} $$Now, evaluate the slope at $(3,3)$:
$$ m = \frac{2(3) - (3)^2}{(3)^2 - 2(3)} = \frac{6-9}{9-6} = \frac{-3}{3} = -1 $$Using the point-slope formula: $y - 3 = -1(x - 3) \implies y = -x + 6$.
Find the equation of the tangent line to the circle $x^2 + y^2 = 25$ at the point $(3, -4)$.
To find the second derivative, $\frac{d^2y}{dx^2}$, we differentiate our expression for $\frac{dy}{dx}$ a second time with respect to $x$.
The Key Step for Simplification: After differentiating, your expression for $\frac{d^2y}{dx^2}$ will contain a $\frac{dy}{dx}$ term. You must substitute the expression you found for the first derivative back into the equation. Then, look for opportunities to use the original relation to simplify even further.
Find $\frac{d^2y}{dx^2}$ for $x^2 + y^2 = 25$.
We start with $\frac{dy}{dx} = -\frac{x}{y}$. Now, differentiate using the Quotient Rule:
$$ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(-\frac{x}{y}\right) = - \frac{(y)(1) - (x)(\frac{dy}{dx})}{y^2} $$Substitute $\frac{dy}{dx} = -x/y$ into the expression:
$$ \frac{d^2y}{dx^2} = - \frac{y - x(-x/y)}{y^2} = - \frac{y + x^2/y}{y^2} $$To simplify the complex fraction, multiply the top and bottom by $y$:
$$ \frac{d^2y}{dx^2} = - \frac{y^2 + x^2}{y^3} $$Now for the final simplification: we know from the original relation that $x^2+y^2=25$. So we substitute that in:
$$ \frac{d^2y}{dx^2} = -\frac{25}{y^3} $$Find the value of $y''$ at the point $(2,1)$ for the ellipse $x^2 + 4y^2 = 8$.
Find $\frac{dy}{dx}$ for the lemniscate given by $(x^2+y^2)^2 = 2(x^2-y^2)$.
Find all points on the curve $x^2+2y^2-x=0$ where the tangent line is vertical.
Find $\frac{dy}{dx}$ for the relation $xe^y = x-y$.
This lecture introduced implicit differentiation, a technique for finding derivatives when $y$ is not explicitly defined as a function of $x$. The method relies on applying the Chain Rule to terms involving $y$ when differentiating with respect to $x$. We used this technique to find the slopes of tangent lines for complex curves and to find the second derivative.
After this lecture, you should be able to: