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.
Why do we multiply by dy/dx? Open for an explanation.
Let’s consider, as an example, the function $e^{x^2}$. When you take its derivative, you of course use the Chain Rule:
\begin{align*}
\dfrac{d}{dx}\left(e^{x^2} \right) &= e^{x^2} \cdot \dfrac{d}{dx}\left(x^2 \right) \\[8px]&= e^{x^2} \cdot (2x)
\end{align*}Similarly, when you take the derivative of, say, $e^{f(x)}$, where $f(x)$ is some function we’re not specifying, you again of course use the Chain Rule:
\begin{align*}
\dfrac{d}{dx}\left(e^{f(x)} \right) &= e^{f(x)} \cdot \dfrac{d}{dx}\left(f(x) \right) \\[8px]&= e^{f(x)} \cdot \dfrac{df}{dx}
\end{align*}
That might look funny, but we don’t know what $\dfrac{df}{dx}$ is, so we’re just leaving it written like that.And also similarly, let’s consider the derivative of, say, $e^{y(x)}$, where we’re writing $y(x)$ to indicate that y is a function of x. We again of course must use the Chain Rule:
\begin{align*}
\dfrac{d}{dx}\left(e^{y(x)} \right) &= e^{y(x)} \cdot \dfrac{d}{dx}\left(y(x) \right) \\[8px]&= e^{y(x)} \cdot \dfrac{dy(x)}{dx}
\end{align*}
Now we don’t usually write $y(x)$, because it’s annoying to write again and again; instead we just write “y”—but always have to keep in mind that really y still depends on x. We thus usually write that last result as$$\dfrac{d}{dx}\left(e^y \right) = e^y \cdot \dfrac{dy}{dx}$$
The key point is that we end up multiplying by $\dfrac{dy}{dx}$, because the function we were taking the derivative of, $e^y$, has a y in it. Similarly, you should see that the following are all correct derivatives for these example functions:
\begin{align*}
\dfrac{d}{dx}[y ] &= 1 \cdot \dfrac{dy}{dx} = \dfrac{dy}{dx}\\[8px]\dfrac{d}{dx}\left[y^5 \right] &= 5y^4 \cdot \dfrac{dy}{dx} \\[8px]\dfrac{d}{dx}[\sin(y) ] &= \cos(y) \cdot \dfrac{dy}{dx} \\[8px]\dfrac{d}{dx}\left[ \cos(y^2) \right] &= -\sin(y^2) \cdot 2y \cdot \dfrac{dy}{dx} \\[8px]\dfrac{d}{dx}[\ln(y)] &= \frac{1}{y} \cdot \dfrac{dy}{dx} \\[8px]\dfrac{d}{dx}\left[\ln(y^5) \right] &= \frac{1}{y^5} \cdot 5y^4 \cdot \dfrac{dy}{dx} = \frac{5}{y} \cdot\dfrac{dy}{dx}
\end{align*}The biggest challenge when learning to do Implicit Differentiation problems is to remember to include this $\dfrac{dy}{dx}$ term when you take the derivative of something that has a y in it. As always, practicing is the way to learn, and you’ll get good practice problems below.
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- 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 y in it, be sure to multiply by $\dfrac{dy}{dx}$ due to the Chain Rule.
- 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*}
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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!)
Insurance against making a mistake
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.)
Show/Hide Step 1
We first make the substitution $y = f(x).$ Remember that this is optional; you can certainly write everything in terms of $y$ and $y’$ if you prefer.
\begin{align*}
y^5 + xy^2 + x &= xy^4 + x^2 + 4y \\[8px] f(x)^5 + xf(x)^2 + x &= xf(x)^4 + x^2 + 4f(x)
\end{align*}
Next we take the derivative of both sides of the equation, remembering to apply the Chain Rule, and also the Product Rule where necessary. We’ll take this opportunity to apply a good book-keeping practice and think about which rule(s) we’ll apply to each term at this outset. You probably won’t ever write each rule as we have here, but you’d be doing the same thing mentally:
\begin{align*}
& \overbrace{f(x)^5}^{\text{Chain}} + \overbrace{xf(x)^2}^{\text{Product and Chain}} + x = \overbrace{xf(x)^4}^{\text{Product and Chain}} + x^2 + \overbrace{4f(x)}^{\text{Chain}} \\[8px] & 5f(x)^4 \cdot f'(x) + f(x)^2 + 2xf(x)f'(x) + 1 = f(x)^4 + 4xf(x)^3f'(x) + 2x + 4f'(x)
\end{align*}
Let’s now un-substitute $f(x) = y$ and $f'(x) = y’,$ and solve for $y’$.
\begin{align*}
& 5y^4\,y’ + y^2 + 2xy\,y’ + 1 = y^4 + 4xy^3\,y’ + 2x + 4y’ \\[8px] & 5y^4\,y’ + 2xy\,y’ – 4xy^3\,y’ -4y’ = y^4 + 2x – y^2 – 1 \\[8px] & y’\left(5y^4 + 2xy – 4xy^3 -4\right) = y^4 + 2x – y^2 – 1 \\[8px] & y’ = \frac{y^4 + 2x – y^2 – 1}{5y^4 + 2xy – 4xy^3 -4} \quad \blacktriangleleft
\end{align*}
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Step 2: Compute the slope of the tangent line at the point point $(-3,-2).$
Show/Hide Step 2
To find the slope of a tangent at our particular point $(-3,-2)$, we evaluate $y’$ with $x=-3$ and $y=-2$:
\begin{align*}
y'(-3,-2) &= \frac{(-2)^4 + 2(-3) – (-2)^2 – 1}{5(-2)^4 + 2(-3)(-2) – 4(-3)(-2)^3 -4} \\[8px] &= \frac{16 – 6 – 4 – 1}{5(16) + 12 – (-12)(-8) -4} \\[8px] &= \frac{5}{80 + 12 – 96 -4} \\[8px] &= -\frac{5}{8} \quad \blacktriangleleft
\end{align*}
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Step 3: Write the equation of the tangent line in point-slope form, $y-y_1 = m(x-x_1).$
Show/Hide Step 3
We now have the slope of the tangent line is $m = -\dfrac{5}{8},$ and we know the line passes through the point $(-3,-2).$ Hence
\begin{align*}
y-y_1 &= m(x-x_1) \\[8px] y-(-2) &= -\frac{5}{8}\Big(x-(-3)\Big) \\[8px] y+2 &= -\frac{5}{8}(x+3) \quad \cmark
\end{align*}
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Practice Problems
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}
Show/Hide Solution
. Take $\dfrac{d}{dx}$ of both sides of the equation. \begin{align*} \dfrac{d}{dx}\left( y^3+2x^2 + y \right) &= \dfrac{d}{dx}\left( 3x^4 \right) \\[8px] \left( 3y^2 \right)\left( \dfrac{dy}{dx} \right) + 4x + \dfrac{dy}{dx} &= 12x^3 \\[8px] \end{align*}
Step 2. Solve for $\dfrac{dy}{dx}$. \begin{align*} 3y^2 \dfrac{dy}{dx} + \dfrac{dy}{dx} &= 12x^3 – 4x \\[8px] \dfrac{dy}{dx} \left(3y^2 + 1 \right) &= 12x^3 – 4x \\[8px] \dfrac{dy}{dx} &= \left( 4x \right)\dfrac{3x^2-1}{1+3y^2} \implies \; \text{ (B) } \; \cmark \end{align*}
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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}
Show/Hide Solution
. Take the derivative of both sides of the equation. \begin{align*} \left[ e^x+ \cos y \right]’ &= \left[ 6y \right]’ \\[8px] e^x – (\sin y) \cdot y’ &= 6 y’ \\[8px] \end{align*}
Step 2. Solve for $y’$. \begin{align*} e^x – y’ \sin y &= 6y’ \\[8px] e^x &= y’ \left( 6 + \sin y \right)\\[8px] y’ &= \dfrac{e^x}{6+ \sin y} \implies \; \text{ (A) } \; \cmark \end{align*}
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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}
Show/Hide Solution
We can rewrite the equation: $x^2 = y^2 \sin y$. Then, \begin{align*} \left[ x^2 \right]’ &= \left[ y^2 \sin y \right]’\\[8px] \left[ x^2 \right]’ &= \left[ y^2 \right]’ \cdot \sin y + y^2 [\sin y]’ \\[8px] 2x &= [2y\, y’] \sin y + y^2 [(\cos y) y’] \\[8px] 2x &= y’ \left( 2y \sin y + y^2 \cos y \right) \\[8px] y’ &= \dfrac{2x}{ 2y \sin y + y^2 \cos y } \implies \; \text{ (B) } \; \cmark \end{align*}
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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}
Show/Hide Solution
We can substitute $y = f(x)$ and determine an expression for $y’$. \begin{align*} x^{4}y^{2} &= 4x^{5} + \sin y \\[8px] \left[ x^{4}y^{2} \right]’ &= \left[ 4x^{5} + \sin y \right]’ \\[8px] 4x^{3}y^2 + x^{4}(2y\,y’) &= 20x^{4} +( \cos y) y’ \\[8px] 4x^{3}y^{2} – 20x^{4} &= y’ \left( \cos y -2yx^{4} \right) \\[8px] y’ &= \dfrac{4x^{3} \left( y^2 – 5x \right)}{ \cos y – 2yx^{4}} \\[8px] f'(x) &= \dfrac{4x^{3} \left[ (f(x))^2 – 5x \right]}{ \cos \left[f(x) \right] – 2f(x)x^{4}} \implies \; \text{ (C) } \; \cmark \end{align*}
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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}
Show/Hide Solution
Step 1: Take $\dfrac{d}{dx}$ of both sides of the equation and then solve for $\dfrac{dg}{dx}$. \begin{align*} \dfrac{d}{dx} \left[ \left(x^2 g(x)\right) + \sin x \right] &= \dfrac{d}{dx}\left[ g(x) \right]^2 \\[8px] \left( 2xg(x) + x^2 \dfrac{dg}{dx} \right) + \cos x &= 2g(x)\dfrac{dg}{dx} \\[8px] \end{align*} \begin{align*} 2g(x)\dfrac{dg}{dx} – x^2 \dfrac{dg}{dx} &= 2xg(x) + \cos x \\[8px] \dfrac{dg}{dx} \left[ 2g(x) – x^2 \right] &= 2x g(x) + \cos x \\[8px] \dfrac{dg}{dx} &= \dfrac{ 2x\, g(x) + \cos x }{ 2g(x) – x^2 } \\[8px] \end{align*} Step 2: We need the value of $g(x)$ at $x=\pi$: \begin{align*} x^2g(x)+ \sin x &= \left[ g(x) \right]^2 \\[8px] \pi ^2 g(\pi) + \sin (\pi) &= [g(\pi)]^2 \\[8px] \pi ^2 g(\pi) + 0 &= [g(\pi)]^2 \\[8px] g(\pi) &= \pi ^2 \\[8px] \end{align*} Step 3: Find $g'(\pi)$. We start with our result from Step 1: \begin{align*} g'(x) = \dfrac{dg}{dx} &= \dfrac{ 2x \,g(x) + \cos x }{ 2g(x) – x^2 } \\[8px] g'(\pi) &= \dfrac{ 2 \pi (\pi ^2) + \cos \pi }{2 \pi ^2 – \pi ^2} \\[8px] &= \dfrac{2 \pi ^{3} -1 }{ \pi ^2 } \\[8px] &= 2 \pi – \dfrac{1}{\pi ^2} \implies \; \text{ (C) } \; \cmark \end{align*}
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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}
Show/Hide Solution
\begin{align*} \dfrac{d}{dx} \left( \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} \right) &= \dfrac{d}{dx}(1) \\[8px] \dfrac{2x}{a^2} + \dfrac{2y}{b^2}\,y’ &= 0 \\[8px] y’ \dfrac{2y}{b^2} &= -\dfrac{2x}{a^2} \\[8px] y’ &= -\dfrac{b^2}{a^2} \dfrac{x}{y} \implies \; \text{ (D) } \; \cmark \end{align*} Bonus: The circle is a special kind of ellipse where $a=b$.
Applying this to our previous answer, we get: \begin{align*} y’ &= -\dfrac{b^2}{a^2} \dfrac{x}{y} \\[8px] y’ &= -\dfrac{a^2}{a^2} \dfrac{x}{y} \\[8px] y’ &= -\dfrac{x}{y} \; \cmark \end{align*}
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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}
Correct
$x=1$ for both values. Use the points $(1,2)$ and $(1,-3)$ when determining $y'$ values.
$x=1$ for both values. Use the points $(1,2)$ and $(1,-3)$ when determining $y'$ values.
$x=1$ for both values. Use the points $(1,2)$ and $(1,-3)$ when determining $y'$ values.
Start by using Implicit Differentiation and solve for $y'$. Substitute $x=1$ into the original equation and note that will yield a quadratic equation in $y$. We need both $y$ values to determine both values of $y'$
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Let us begin as we normally would, finding $y’$ using Implicit Differentiation. \begin{align*} \dfrac{d}{dx} \left( x^2 + xy + y^2 \right) &= \dfrac{d}{dx} (7) \\[8px] 2x + (y + xy’) + 2yy’ &= 0 \\[8px] \end{align*} Solving for $y’$, \begin{align*} xy’ + 2yy’ &= – \left( 2x+y \right) \\[8px] y’ \left( x+2y \right) &= – \left( 2x+y \right) \\[8px] y’ &= -\dfrac{ 2x+y }{ x+2y } \\[8px] \end{align*} As we can see from the graph the curve has two $y$-values at $x=1$. We need to determine each using the original equation:
\begin{align*} x^2 + xy +y^2 &= 7 \\[8px] 1 + y + y^2 &= 7 \\[8px] y^2 + y – 6 &= 0 \\[8px] (y-2)(y+3) &= 0 \\[8px] y = 2 &\text{ and } y = -3 \; \blacktriangleleft \end{align*} The two points we must consider are thus $(1, 2)$ and $(1, -3)$.
Now recall our result from above, \[y’ = -\dfrac{ 2x+y }{ x+2y }\] $y’$ for the point in the first quadrant, $(1,2),$ is given by: \begin{align*} y’ &= -\dfrac{ 2(1)+(2) }{ 1+2(2) } \\[8px] &= – \dfrac{4}{5} \\[8px] \end{align*} $y’$ for the point in the fourth quadrant, $(1,-3),$ is given by: \begin{align*} y’ &= -\dfrac{ 2(1)+(-3) }{ (1)+2(-3) } \\[8px] &= – \dfrac{-1}{-5} \\[8px] &= – \dfrac{1}{5} \end{align*} Hence our two values are \[-\dfrac{4}{5}, \; -\dfrac{1}{5} \implies \; \text{ (A) } \; \cmark \]
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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}
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Step 1: Take the derivative of both sides of the equation and then solve for $y’$.
\begin{align*} \left[ \dfrac{x^2}{4} + \dfrac{y^2}{9} \right]’ &= [1]’ \\[8px] \dfrac{2x}{4} + \dfrac{2y\,y’}{9} &= 0 \\[8px] \dfrac{2y\,y’}{9} &= -\dfrac{2x}{4} \\[8px] \dfrac{y\,y’}{9} &= -\dfrac{x}{4} \\[8px] y’ &= -\dfrac{9}{4}\dfrac{x}{y} \\[8px] \end{align*} Step 2: Take the derivative of both sides again and solve for $y”$.
To avoid having to use the Quotient rule, which just makes life much easier, rewrite the preceding line as $y’\,y = -\dfrac{9}{4}x$: \begin{align*} \dfrac{d}{dx} \left( y’y \right) &= -\dfrac{9}{4} \dfrac{d}{dx} (x) \\[8px] y”y + y’y’ &= -\dfrac{9}{4} \\[8px] y”y &= -\dfrac{9}{4} – (y’)^2 \\[8px] y” &= -\dfrac{ \frac{9}{4} + (y’)^2 }{ y } \\[8px] \end{align*} Recall from above that $y’ = -\dfrac{9}{4}\dfrac{x}{y}$: \begin{align*} y” &= – \dfrac{ \dfrac{9}{4} + \left( -\dfrac{9}{4}\dfrac{x}{y} \right)^2 }{y} \\[8px] &= – \dfrac{\dfrac{9}{4} \left(1+ \dfrac{9}{4} \dfrac{x^2}{y^2}\right)}{y} \\[8px] &= -\dfrac{9}{4} \dfrac{y^2 + \frac{9}{4}x^2}{y^3} \implies \; \text{ (D) } \; \cmark \end{align*}
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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}
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Let’s first find the value of $y$ at $x=0$: \begin{align*} xy + e^y &= e \\[8px] (0)(y) + e^y &= e \\[8px] e^y &= e \\[8px] y &= 1 \\[8px] \end{align*} So our point of interest is $(0, 1).$
Next, we’ll take $\dfrac{d}{dx}$ of both sides of the equation and solve for $y’$. \begin{align*} \dfrac{d}{dx} \left( xy + e^y \right) &= \dfrac{d}{dx}(e) \\[8px] y+xy’ + e^y y’ &= 0 \\[8px] y'(x+e^y) &= -y \\[8px] y’ &= -\dfrac{y}{x+e^y} \\[8px] \end{align*} Let’s take the derivative again to find $y”$: \begin{align*} y’ \left( x+e^y \right) &= -y \\[8px] \dfrac{d}{dx} \left[ y’ \left( x+e^y \right) \right] &= – \dfrac{d}{dx} (y) \\[8px] y”\left( x+e^y \right) + y’ \left( 1 + e^y y’ \right) &= -y’ \\[8px] y” &= (y’)\dfrac{ -1 – \left( 1+e^y(y’) \right) }{ \left( x+e^y \right) } \\[8px] \end{align*} Note that we found above $y’ = -\dfrac{y}{x+e^y} $. Hence at $(0,1)$, $y’= -\dfrac{1}{e}$.
We finally find $y”$ at the point $(0, 1)$: \begin{align*} y” &= (y’)\dfrac{ -1 – \left( 1+e^y(y’) \right) }{ \left( x+e^y \right) } \\[8px] \left. y” \right|_{(0,\,1)} &= \left( – \dfrac{1}{e}\right) \dfrac{-2 – e \left( \frac{-1}{e} \right)}{0+e}\\[8px] &= \left( – \dfrac{1}{e}\right)\dfrac{-2+1}{e} \\[8px] &= \dfrac{1}{e^2} \implies \; \text{ (B) } \; \cmark \end{align*}
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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}
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To write the equation of a line, we need to know its slope at the point of interest, $\left( \pi, \dfrac{ \pi }{2} \right)$.
Let’s first find the slope, $y’=\dfrac{dy}{dx}$, at $\left( \pi, \dfrac{ \pi }{2} \right)$: \begin{align*} \left[ y \sin x \right]’ &= \left[ x \cos y \right]’ \\[8px] y’ \sin x + y \cos x &= \cos y – x (\sin y) \cdot y’ \\[8px] y’ \sin x + xy’ \sin y &= \cos y – y \cos x \\[8px] y’ \left( \sin x + x \sin y \right) &= \cos y – y \cos x \\[8px] y’ &= \dfrac{ \cos y – y \cos x }{ \sin x + x \sin y } \\[8px] \end{align*} Evaluating $y’$ at the point $\left( \pi, \dfrac{ \pi }{2} \right)$: \begin{align*} y’ &= \dfrac{ \cos \left( \dfrac{ \pi }{2} \right) – \left( \dfrac{ \pi }{2} \right) \cos \left(\pi \right) }{ \sin (\pi) + \pi \sin \left( \dfrac{\pi}{2} \right) } \\[8px] &= \dfrac{ 0 – \left( \dfrac{ \pi }{2} \right) (-1) }{ 0 + \left(\pi \right) (1) } \\[8px] &= \dfrac{\dfrac{\pi}{2}}{\pi} \\[8px] &= \dfrac{1}{2} \\[8px] \end{align*} We write the equation for the tangent line with slope $\dfrac{1}{2}$ at the point $( \pi, \dfrac{ \pi }{2} )$ in Point-Slope form and then simplify: \begin{align*} y – \dfrac{ \pi }{2} &= \dfrac{1}{2} \left( x – \pi \right) \\[8px] y &= \dfrac{1}{2}x – \dfrac{\pi}{2} + \dfrac{ \pi }{2} \\[8px] y &= \dfrac{1}{2}x \implies \; \text{ (A) } \; \cmark \end{align*}
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The Upshot
- 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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