On the preceding screen we illustrated what an “implicit function” (which really means “a function defined implicitly”) is. As promised, let’s now see how we can easily apply our Calculus tools to take the derivative of such a function, to find $\dfrac{dy}{dx}$ or $y'(x).$ We’ll of course work some examples, and provide practice problems so you can develop your skills.

For most students, the process of implicit differentiation is straightforward, building on all of the other “taking derivatives” work we’ve done. There is one new piece you just have to get used to, which is using the Chain Rule in a particular way. It’sin the first bullet point of Step 1 in our Problem Solving Strategy:

PROBLEM SOLVING STRATEGY: Implicit Differentiation

There are two basic steps to solve implicit differentiation problems:

- Take the derivative $\dfrac{d}{dx}$ of both sides of the equation.
- Use your usual Rules of Differentiation, with one addition: When you take the derivative of a term with a
*y*in it, be sure to multiply by $\dfrac{dy}{dx}$ due to the Chain Rule. - Remember that $\dfrac{d}{dx}\text{(constant)}= 0$ . (It’s a common error to forget that when doing these problems, even among experts.)

- Use your usual Rules of Differentiation, with one addition: When you take the derivative of a term with a
- Solve for $\dfrac{dy}{dx}$.
- Collect all terms with $\dfrac{dy}{dx}$ in them on the left side of the equation, all other terms on the right.
- Factor and divide as necessary to solve for $\dfrac{dy}{dx}$.

*Note*: You may use $y’$ instead of $\dfrac{dy}{dx}$. They are interchangeable:

$$y’ = \dfrac{dy}{dx}$$

Let’s consider an Example to illustrate, returning again to our unit circle.

Example 1: $x^2 + y^2 = 1$

**(a)** Given $x^2 + y^2 = 1,$ find $\dfrac{dy}{dx}.$

**(b)** Find the slope of the tangent lines to the circle at $x = 0.5$.

*Solution.*

**(a)** Let’s find $\dfrac{dy}{dx}$:

__Step 1__. Take the derivative $\dfrac{d}{dx}$ of both sides of the equation. Remember the Chain Rule as applied to *y.*

\begin{align*}

\dfrac{d}{dx}\left[ x^2 \right] + \dfrac{d}{dx}\left[y^2 \right] &= \dfrac{d}{dx}[1] \\[8px]
2x + 2y \cdot \dfrac{dy}{dx} &= 0

\end{align*}

__Step 2__. Solve for $\dfrac{dy}{dx}.$

\begin{align*}

2y \cdot \dfrac{dy}{dx} &= -2x \\[8px]
\dfrac{dy}{dx} &= \frac{-x}{y} \quad \cmark

\end{align*}

(b) To find the slope of the tangent lines to the circle at $x = 0.5,$ notice first that the equation for $\dfrac{dy}{dx}$ that we just found has a *y* in it. Hence we need the *y*-values of the circle at $x = 0.5.$

Solving the circle equation for *y* gives

\[y = \pm \sqrt{1 – x^2}\]
Hence at $x = 1,$ the *y*-values are

\[ y_1 = \sqrt{1 – (0.5)^2} = 0.866 \qquad \text{and} \qquad y_2 = -\sqrt{1 – (0.5)^2} = -0.866 \]
We found in part (a) that $\dfrac{dy}{dx} = \dfrac{-x}{y}.$ Hence:

• At $(0.5, 0.866),$ the slope is

\[ \left. \dfrac{dy}{dx} \right|_{(0.5, 0.866)} = \frac{-0.5}{0.866} = -0.577 \] • And at $(0.5, -0.866),$ the slope is

\[ \left. \dfrac{dy}{dx} \right|_{(0.5, -0.866)} = \frac{-0.5}{-0.866} = 0.577 \]

Because we can solve the circle equation for *y,* we don’t *have* to use implicit differentiation to find $\dfrac{dy}{dx},$ as illustrated in the box below. However, using implicit differentiation results in a single equation that works for all $(x, y)$ on the circle, whereas completing the calculation without implicit differentiation requires treating the top and bottom halves of the circle separately.

Show/Hide Example 1 result without Implicit Differentiation

We know we can represent the top and bottom halves of the circle with the equations

\[y_1(x) = \sqrt{1-x^2} \quad \text{and} \quad y_2(x) = -\sqrt{1-x^2}\] Let’s find the derivative of each, starting with $y_1(x):$

\begin{align*}

\dfrac{dy_1}{dx} &= \dfrac{d}{dx}\left(1 – x^2 \right)^{1/2} \\[8px] &= \dfrac{1}{2}\left(1 – x^2 \right)^{-1/2}\cdot (-2x) \\[8px] &= -\frac{x}{\sqrt{1-x^2}} \quad \cmark

\end{align*}

While this result might initially look different than what we found in Example 1, remember that $y_1 = \sqrt{1-x^2},$ and so we can rewrite the preceding line as

\[\dfrac{dy_1}{dx} = \frac{-x}{y_1}\] The calculation for $\dfrac{dy_2}{dx}$ is the same, except for the negative sign in front, so

\begin{align*}

\dfrac{dy_2}{dx} &= \frac{x}{\sqrt{1-x^2}} \quad \cmark \\[8px] &= \frac{-x}{y_2}

\end{align*}

[collapse]

Furthermore, in situations where we cannot solve the equation for an explicit form of $y(x) =\, \rule{0.6cm}{0.15mm},$ we have no choice but to use implicit differentiation. (And besides, it’s so much easier, there’s no reason you would ever want to *not* use it!)

*The* most common error students make, especially on exams, is to forget to use the Chain Rule on one or more terms involving *y* when taking the derivative. A simple way to help avoid this error is to add a step to the beginning and end of the procedure, and replace *y* with $f(x).$ That is:

- Replace
*y*with $f(x)$ in the equation. - Take the derivative $\dfrac{d}{dx}$ of both sides of the equation. Remember the Chain Rule, so every term in the original equation that has an $f(x)$ will have now contain $\dfrac{df}{dx}.$
- Solve for $\dfrac{df}{dx}.$
- If you’d like or if required, substitute back $f(x) = y.$

The next example illustrates. We’ll also use prime notation for the derivative to show how that works as well.

Example 2: $x^2 + y = 2y^2 + 1$

Use implicit differentiation to find $y’$ given $x^2 + y = 2y^2 + 1.$

*Solution.*

__Step 1__. Replace *y* with $f(x)$ in the equation.

\[x^2 + f(x) = 2f(x)^2 + 1\] __Step 2__. Take the derivative of both sides of the equation with respect to *x.*

\[2x + f'(x) = 4f(x)\cdot f'(x)\] __Step 3__. Solve for f'(x).

\begin{align*}

2x + f'(x) &= 4f(x)\cdot f'(x) \\[8px]
f'(x) – 4f(x)\cdot f'(x) &= -2x \\[8px]
f'(x) \cdot (1-4f(x)) &= -2x \\[8px]
f'(x) &= \frac{-2x}{1-4f(x)} \\[8px]
f'(x) &= \frac{2x}{4f(x)-1}

\end{align*}

__Step 4__. Especially since the question asked us to find $y’,$ we’ll substitute back $f(x) = y$ and $f'(x) = y’.$

\[y’ = \frac{2x}{4y-1} \quad \cmark \]

If as you’re practicing you find yourself ever missing the Chain Rule term $\dfrac{dy}{dx}$ or $y’,$ we strongly suggest using the $y = f(x)$ trick illustrated in Example 2. This simple move seems to help cue students to correctly apply the Chain Rule in high-stakes, exam situations.

Let’s work through a Scaffolded Problem, where you can check your work at each key step. The question may appear intimidating at first, but the power of Implicit Differentiation is that it makes taking derivatives of even super-complicated-looking equations straightforward. And once that part is done, you’re left with a standard “write the equation of a tangent line” problem, as you’ll see.

Scaffolded Problem #1: Fifth degree polynomial

Find the equation of the line tangent to the curve

\[y^5 + xy^2 + x = xy^4 + x^2 + 4y\]
at the point $(-3, -2)$.

*Solution.*

**Step 1**: Use implicit differentiation to find $y’.$

(Note that in our solution, we will use the substitution $y = f(x)$ to make sure we don’t forget to apply the Chain Rule.)

Time to practice! These problems can be kinda fun once you get the hang of them, and you’ll probably find that after a few you have the routine down rather solidly. (But more practice never hurts!)

Practice Problem #1

Use implicit differentiation to find $\dfrac{dy}{dx}$ given $y^3+2x^2+y=3x^4$.
\begin{array}{lll} \text{(A) }\dfrac{4x}{y} \left( \dfrac{3x^2-1}{1+3y} \right) && \text{(B) } \left( 4x \right) \dfrac{3x^2-1}{1+3y^2} && \text{(C) } \dfrac{4x}{3y^2} \left( 3x^2-1 \right) \end{array}
\begin{array}{lll}\text{(D) } \dfrac{4x}{y} \left(\dfrac{3x^2-1}{y^2+1} \right) && \text{(E) None of these}
\end{array}

Practice Problem #2

Use implicit differentiation to find $y'$ given $e^x+ \cos y = 6y$.
\begin{array}{lllll} \text{(A) } \dfrac{e^x}{6+ \sin y} && \text{(B) } \sin ^{-1} \left( e^x - 6 \right) && \text{(C) } \cos ^{-1} \left( 6-e^x \right) \end{array}
\begin{array}{lll} \text{(D) } \dfrac{6-e^x}{ \cos y} && \text{(E) None of these}
\end{array}

Practice Problem #3

Use implicit differentiation to find $y'$ given $ \dfrac{x^2}{y^2} = \sin y $.
\begin{array}{lll} \text{(A) } y' = \dfrac{x}{ \cos y + \frac{2x^2}{y} } && \text{(B) } y' = \dfrac{2x}{2y \sin y + y^2 \cos y} && \text{(C) } y' = \dfrac{x}{ \cos y + 2 \sin y }\end{array}
\begin{array}{lll}\text{(D) } y' = \dfrac{x}{ \cos y + \frac{x^2}{y} } && \text{(E) None of these}
\end{array}

Practice Problem #4

Given $x^{4} \left[ f(x) \right]^2 = 4x^{5} + \sin \left[ f(x) \right]$, find an expression for $f'(x)$.
\begin{array}{lll} \text{(A) }f'(x) = \dfrac{x^{3} \left[ (f(x))^2 + 5x \right]}{ \cos \left[f(x) \right] - 2f(x)x^{4}} && \text{(B) }f'(x) = \dfrac{4x \left[ (f(x))^2 - 5x \right]}{ \cos \left[f(x) \right] - 2f(x)x^{4}} \end{array}
\begin{array}{lll}\text{(C) } f'(x)=\dfrac{4x^3 \left[ \left( f(x) \right)^2 -5x \right]}{ \cos \left[ f(x) \right] - 2x^{4}f(x)}
&&\text{(D) } f'(x) = \dfrac{4x^{3} \left[ (f(x))^2 - x \right]}{ \cos \left[f(x) \right] - f(x)x^{4}} && \text{(E) None of these} \end{array}

Practice Problem #5

Given $x^2g(x)+ \sin x = \left[ g(x) \right]^2$ find $g'(\pi)$.
\begin{array}{lllll} \text{(A) }\pi ^2 -1 && \text{(B) }1- \dfrac{1}{\pi ^2} && \text{(C) } 2 \pi - \dfrac{1}{ \pi ^{2} } && \text{(D) } \pi ^2 - \dfrac{1}{2 \pi} && \text{(E) None of these}
\end{array}

Practice Problem #6

Consider an ellipse, $ \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1.$ Find the slope of the curve $y'$ at the point $(x,y).$

Bonus: Show that when $a = b,$ the curve's slope reduces to $y'=-\dfrac{x}{y}$ as we found above for the case of a circle. \begin{array}{lllll} \text{(A) }-\dfrac{a^2}{b^2} \dfrac{x}{y} && \text{(B) }-\dfrac{a}{b} \dfrac{x}{y} && \text{(C) } -\dfrac{b}{a} \dfrac{x}{y} && \text{(D) }-\dfrac{b^2}{a^2} \dfrac{x}{y} && \text{(E) }-\dfrac{a^3}{b^3}\dfrac{x}{y} \end{array}

Bonus: Show that when $a = b,$ the curve's slope reduces to $y'=-\dfrac{x}{y}$ as we found above for the case of a circle. \begin{array}{lllll} \text{(A) }-\dfrac{a^2}{b^2} \dfrac{x}{y} && \text{(B) }-\dfrac{a}{b} \dfrac{x}{y} && \text{(C) } -\dfrac{b}{a} \dfrac{x}{y} && \text{(D) }-\dfrac{b^2}{a^2} \dfrac{x}{y} && \text{(E) }-\dfrac{a^3}{b^3}\dfrac{x}{y} \end{array}

Practice Problem #7

Consider the relationship $x^2 + xy +y^2=7,$ which is shown in the graph. Find both values of $y'$ for the curve at $x=1$.

\begin{array}{lllll} \text{(A) }-\dfrac{4}{5},\; -\dfrac{1}{5} && \text{(B) }\dfrac{5}{3},\;5 && \text{(C) } \dfrac{5}{3},\;-\dfrac{1}{5} && \text{(D) } -\dfrac{4}{5},\;5 && \text{(E) None of these}

\end{array}

Practice Problem #8

Find $y''$ for the ellipse given by $\dfrac{x^2}{4} + \dfrac{y^2}{9}=1$.
\begin{array}{lll} \text{(A) }-\dfrac{9}{4}\dfrac{x}{y} && \text{(B) }-\dfrac{9}{4}\left[ \dfrac{ y-xy' }{y^2} \right] && \text{(C) }-\dfrac{9}{4} \end{array}
\begin{array}{lll}\text{(D) }-\dfrac{9}{4} \left[ \dfrac{y^2 + \frac{9}{4}x^2}{y^3} \right] && \text{(E) None of these}
\end{array}

Practice Problem #9

If $xy + e^y = e$, find $y''$ at $x=0$.
\begin{array}{lllll} \text{(A) }\dfrac{2}{e}\left( 2-\dfrac{1}{e} \right) && \text{(B) }\dfrac{1}{e^2} && \text{(C) }\dfrac{1}{e} && \text{(D) } \dfrac{2}{e}\left( 1-\dfrac{1}{e} \right) && \text{(E) }\dfrac{2}{e^2}
\end{array}

Practice Problem #10

Find the equation of the line tangent to the graph of $y \sin x = x \cos y$ at the point $( \pi, \dfrac{ \pi }{2} )$.
\begin{array}{lllll} \text{(A) }\dfrac{1}{2}x && \text{(B) }\dfrac{\pi}{2}\left( x-1 \right) && \text{(C) }\pi \left( \dfrac{x}{2}-1 \right) && \text{(D) }\dfrac{\pi}{2}x && \text{(E) }\dfrac{1}{2}\left( x+1 \right)
\end{array}

- Using implicit differentiation to find the derivative of an implicitly defined function is straightforward:

__Step 1__: Take the derivative $\dfrac{d}{dx}$ of both sides of the equation. The one thing you must be careful about: Remember the Chain Rule! Any term that includes a*y*with result in a Chain Rule term $\dfrac{dy}{dx}.$

__Step 2__: Solve for $\dfrac{dy}{dx}$.

Do you have an implicit differentiation question you’re working on and could use some help with? Or any other comments or questions about what’s on this screen? Please write in the Comments below, or (better yet) on our Forum, where it’s much easier to write math.

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