7.9 Implicit and parametric differentiation — Learn

Prerequisites: the chain rule and the product rule from Section 7.7; differentiating $\sin t$ and $\cos t$; the equation of a circle from Chapter 3.

By the end of this section you can: differentiate a relation written implicitly in $x$ and $y$; find $\dfrac{dy}{dx}$ from a relation such as $x^2 + y^2 = 25$; find the equation of a tangent to a curve given implicitly; and differentiate a curve given parametrically using $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$.

1. Differentiating an implicit relation

So far every curve has been written as $y = $ something. Some curves are easier to write as a relation between $x$ and $y$, such as $x^2 + y^2 = 25$.

When $y$ depends on $x$, differentiate both sides with respect to $x$. Every time you differentiate a $y$-term, the chain rule adds a factor of $\dfrac{dy}{dx}$.

The derivative of $x^2$ is $2x$, as usual.

Example 1. Find $\dfrac{dy}{dx}$ for the circle $x^2 + y^2 = 25$.

Differentiate both sides:

Solve for the derivative:

The gradient depends on both coordinates, not on $x$ alone.

Example 2. Find $\dfrac{dy}{dx}$ for $x^2 + 3xy - y^2 = 5$.

The middle term is a product, so it needs the product rule. Differentiate each term with respect to $x$:

Collect the $\dfrac{dy}{dx}$ terms on one side:

Therefore

Example 3. Find the equation of the tangent to $x^2 + y^2 = 25$ at the point $(3, 4)$.

From Example 1, $\dfrac{dy}{dx} = -\dfrac{x}{y}$. At $(3, 4)$ the gradient is

The tangent through $(3, 4)$ is

2. Differentiating a parametric curve

Sometimes $x$ and $y$ are each written in terms of a third variable, called a parameter. For example,

The gradient is still $\dfrac{dy}{dx}$, and the chain rule connects the three derivatives:

Example 4. Find $\dfrac{dy}{dx}$ for $x = t^2$, $y = t^3$.

Differentiate each with respect to $t$:

Then

Example 5. Find $\dfrac{dy}{dx}$ for $x = 2\cos t$, $y = 3\sin t$.

so

3. Try it: the implicit tangent explorer

Predict first. The figure shows the circle $x^2 + y^2 = 25$ with a point you can slide around it. Before you touch the slider, predict the gradient of the tangent at $(5, 0)$ and at $(0, 5)$, and predict the sign of the gradient in the first quadrant. Write your predictions down.

Test. Slide the point around the circle and watch the tangent line and the readout. Compare the gradient with $-\dfrac{x}{y}$. Check the two points where the tangent is vertical, and explain what the formula does there.

Explain. Write two sentences. The first explains why $\dfrac{dy}{dx}$ depends on both $x$ and $y$ for this curve. The second explains why the tangent is vertical at the points $(5, 0)$ and $(-5, 0)$.

Watch out. Three mistakes are common with these methods.

Forgetting the chain rule on a $y$-term. The derivative of $y^2$ is $2y\dfrac{dy}{dx}$, not $2y$. Every $y$ is a function of $x$ when you differentiate with respect to $x$.

Leaving out the product rule on $xy$. The derivative of $xy$ is $y + x\dfrac{dy}{dx}$. Treating it as $1 \times \dfrac{dy}{dx}$ loses a term.

Inverting the parametric derivative. The gradient is $\dfrac{dy/dt}{dx/dt}$, with $\dfrac{dy}{dt}$ on top. Writing $\dfrac{dx/dt}{dy/dt}$ gives the reciprocal, which is the gradient of the inverse relation, not of the curve. Re-check: for a rising curve both derivatives have the same sign, so the gradient must be positive.

You should now be able to

• differentiate both sides of an implicit relation;
• apply the chain rule to $y$-terms and the product rule to $xy$;
• find $\dfrac{dy}{dx}$ from a relation and evaluate it at a point;
• find the equation of a tangent to a curve given implicitly;
• find $\dfrac{dy}{dx}$ for a curve given parametrically;
• simplify a parametric derivative by cancelling the common factor.