Implicit Differentiation

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}\left(y \right) &= 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}\left( \sin(y) \right) &= \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}\left(\ln(y) \right) &= \frac{1}{y} \cdot \dfrac{dy}{dx} \\[8px]\dfrac{d}{dx}\left(\ln(y^5) \right) &= \frac{1}{y^5} \cdot \left(5y^4 \right) \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.)
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:

$$y’ = \dfrac{dy}{dx}$$

Question 1: $x^2 - 2y^2 = 4$

Use implicit differentiation to find $\dfrac{dy}{dx}$ given $x^2 - 2y^2 = 4$.

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Answer: $\dfrac{dy}{dx} = \dfrac{x}{2y}$

Step 1: $\dfrac{d}{dx}$ both sides of the equation. \begin{align*} \frac{d}{dx} \left(x^2 – 2y^2\right) &= \frac{d}{dx}(4) \\[8px] \dfrac{d}{dx}\left(x^2 \right) – 2 \dfrac{d}{dx}\left( y^2\right) &= 0 \\[8px] 2x – 2 \left( 2 y \dfrac{dy}{dx}\right) &= 0 \\[8px] 2x -4y \, \frac{dy}{dx} &= 0 \\[8px] \end{align*} Step 2: Solve for $\dfrac{dy}{dx}$. \begin{align*} \phantom{\dfrac{d}{dx}\left(x^2 \right) – 2 \dfrac{d}{dx}\left( y^2\right)}\\ -4y \, \frac{dy}{dx} &= -2x \\[8px] \frac{dy}{dx} &= \frac{-2x}{-4y} \\[8px] &= \frac{x}{2y} \quad \cmark \end{align*}

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Question 2: $2\sin(x) \cos(y) = 5$

Use implicit differentiation to find $\dfrac{dy}{dx}$ given $2 \sin x \, \cos y = 0.4$.

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Answer: $\dfrac{dy}{dx} = \cot x \, \cot y$

Step 1: $\dfrac{d}{dx}$ both sides of the equation. \begin{align*} \frac{d}{dx} \left[2 \sin x \, \cos y\right] &= \frac{d}{dx}(0.4) \\[8px] 2\left(\frac{d}{dx} \sin x \right)(\cos y) + 2 \sin x \left(\frac{d}{dx} \cos y \right) &= 0 \\[8px] 2 \cos x \, \cos y + 2\sin x \left( -\sin y \cdot \frac{dy}{dx} \right) &= 0 \end{align*} Step 2: Solve for $\dfrac{dy}{dx}$. \begin{align*} \phantom{2\left(\frac{d}{dx} \sin x \right)(\cos y) + 2 \sin x \left(\frac{d}{dx} \cos y \right)} \\ -2\sin x \, \sin y \, \frac{dy}{dx} &= – 2\cos x \, \cos y \\[8px] \frac{dy}{dx} &= \frac{ 2\cos x \, \cos y}{2\sin x \, \sin y} \\[8px] &= \cot x \, \cot y \end{align*}

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Question 3: $e^{(x/y)} = 2x - y$

Use implicit differentiation to find $\dfrac{dy}{dx}$ given $e^{x/y} = 2x - y$.
Note: Don't spend time simplifying your final expression. Once you've isolated $\dfrac{dy}{dx}$, stop.

Show/Hide Solution

Answer: $\dfrac{dy}{dx} = \dfrac{2 – \frac{e^{x/y}}{y}}{1 – \frac{xe^{x/y}}{y^2}} $

Step 1: $\dfrac{d}{dx}$ both sides of the equation. \begin{align*} \dfrac{d}{dx}\left( e^{x/y}\right) &= \dfrac{d}{dx}(2x – y) \\[8px] e^{x/y} \cdot \dfrac{d}{dx} \left(\frac{x}{y} \right) &= 2 – \dfrac{dy}{dx} \\[8px] e^{x/y} \left(\frac{\left( \dfrac{d}{dx}x\right)y – x \left(\dfrac{d}{dx}y \right)}{y^2} \right) &= 2 – \dfrac{dy}{dx} \\[8px] e^{x/y} \left(\frac{1\cdot y – x \dfrac{dy}{dx} }{y^2} \right) &= 2 – \dfrac{dy}{dx} \\[8px] e^{x/y}\left(\frac{1}{y} – \frac{x}{y^2}\dfrac{dy}{dx} \right)&= 2 – \dfrac{dy}{dx} \end{align*} Step 2: Solve for $\dfrac{dy}{dx}$. \begin{align*} \phantom{e^{x/y} \left(\frac{\left( \dfrac{d}{dx}x\right)y – x \left(\dfrac{d}{dx}y \right)}{y^2} \right)}\\ – \frac{x}{y^2}e^{x/y}\dfrac{dy}{dx} + \dfrac{dy}{dx} &= 2 – \frac{e^{x/y} }{y} \\[8px] \left(1 – \frac{x}{y^2}e^{x/y}\right) \dfrac{dy}{dx} &= 2 – \frac{e^{x/y} }{y} \\[8px] \dfrac{dy}{dx} &= \frac{2 – \frac{e^{x/y} }{y}}{1 – \frac{x}{y^2}e^{x/y}} \quad \cmark \end{align*}

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Question 4: $(x+y)^{1/2} + (xy)^{1/2} = 3$

Use implicit differentiation to find $\dfrac{dy}{dx}$ given $\sqrt{x+ y} + \sqrt{xy} = 3$.
Note: Don't spend time simplifying your final expression. Once you've isolated $\dfrac{dy}{dx}$, stop.

Show/Hide Solution

Answer: $\dfrac{dy}{dx} = \dfrac{-\frac{1}{2} \frac{1}{\sqrt{x+ y}} – \frac{1}{2}\frac{y}{\sqrt{xy}}}{ \frac{1}{2} \frac{1}{\sqrt{x+ y}} + \frac{1}{2}\frac{x}{\sqrt{xy}}}$

Step 1: $\dfrac{d}{dx}$ both sides of the equation. \begin{align*} \frac{d}{dx} \left[\sqrt{x+ y} + \sqrt{xy} \right] &= \frac{d}{dx}(3) \\[8px] \frac{d}{dx} \left[\sqrt{x+ y}\right] + \frac{d}{dx}\left[\sqrt{xy} \right] &= 0 \\[8px] \frac{1}{2} \frac{1}{\sqrt{x+ y}} \cdot \frac{d}{dx}(x+y) + \frac{1}{2}\frac{1}{\sqrt{xy}} \cdot \frac{d}{dx}(xy) &= 0 \\[8px] \frac{1}{2} \frac{1}{\sqrt{x+ y}} \left(1 + \frac{dy}{dx} \right) + \frac{1}{2}\frac{1}{\sqrt{xy}} \left(1 \cdot y + x\frac{dy}{dx} \right) &= 0 \end{align*} Step 2: Solve for $\dfrac{dy}{dx}$. \begin{align*} \phantom{\frac{1}{2} \frac{1}{\sqrt{x+ y}} \left(1 + \frac{dy}{dx} \right) + \frac{1}{2}\frac{1}{\sqrt{xy}} \left(1 \cdot y + x\frac{dy}{dx} \right)}\\ \frac{1}{2} \frac{1}{\sqrt{x+ y}}\frac{dy}{dx} + \frac{1}{2}\frac{x}{\sqrt{xy}}\frac{dy}{dx} &= -\frac{1}{2} \frac{1}{\sqrt{x+ y}} – \frac{1}{2}\frac{y}{\sqrt{xy}} \\[12px] \left( \frac{1}{2} \frac{1}{\sqrt{x+ y}} + \frac{1}{2}\frac{x}{\sqrt{xy}}\right) \frac{dy}{dx} &= -\frac{1}{2} \frac{1}{\sqrt{x+ y}} – \frac{1}{2}\frac{y}{\sqrt{xy}} \\[12px] \frac{dy}{dx} &= \frac{-\frac{1}{2} \frac{1}{\sqrt{x+ y}} – \frac{1}{2}\frac{y}{\sqrt{xy}}}{ \frac{1}{2} \frac{1}{\sqrt{x+ y}} + \frac{1}{2}\frac{x}{\sqrt{xy}}} \quad \cmark \end{align*}

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Question 5: Tangent and normal lines

Consider the relation $y^2x^2 = 256$.

(a) Find the points where the tangent line to the curve is parallel to the line $y = 3 -4x$.

(b) Find the points where the tangent line to the curve is perpendicular to the line $y = x$.

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Solution SummarySolution (a) DetailSolution (b) Detail

(a) (-2, -8), (-2, 8) (2, -8) and (2, 8)

(b) (-4, -4), (-4, 4), (4, -4), and (4, 4)

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The line given, $y = 3 – 4x$, is in slope-intercept form ($y = b + mx$), and so we know immediately that its slope is $m = -4$. We therefore want to find points on the curve that have tangent lines that also have slope, which means points on the curve where $y’ = -4$.

\begin{align*} \frac{d}{dx} \left[ y^2x^2 \right] &= \frac{d}{dx}(256) \\ \\ \left(\frac{d}{dx}y^2 \right)x^2 + y^2\left(\frac{d}{dx}x^2 \right) &= 0 \\ \\ \left( 2y y’ \right) x^2 + y^2 (2x) &= 0 \\ \\ \left( 2y x^2 \right)y’ &= -2xy^2 \\ \\ y’ &= -\frac{2xy^2}{2yx^2} = -\frac{y}{x} \end{align*} Now, we want $y’ = -4$, so: \begin{align*} y’ = -\frac{y}{x} &= -4 \\ \\ y&= 4x \end{align*} So we know that the curve has slope $y’ = -4$ when $y = 4x$. To find the points on the curve when this is true, we substitute this requirement into our original relation:

\begin{align*} y^2x^2 &= 256 \\ \\ (4x)^2 x^2 &= 256 \\ \\ 16 x^4 &= 256 \\ \\ x^4 &= 16 \\ \\ x &= \pm 2 \end{align*} We know from our $y’$-requirement above we must have $y = 4x$, and so

$$ y = 4x = \pm 8$$ We can double check the $y$-value by remembering the solutions $x = \pm 2$ and $y = \pm 8$ must satisfy our original relation:

$$y^2x^2 = (\pm 8)^2(\pm 2)^2 = (64)(4) = 256 \checkmark$$ Hence the requested points on the curve are (-2, -8), (-2, 8) (2, -8) and (2, 8). $\quad \cmark$

The line given, $y = x$, is in slope-intercept form ($y = b + mx$), and so we know immediately that its slope is $m_1 = 1$. We want to find points on the curve that have tangent lines that are perpendicular to the that line, and so that have slope

$$m_2 = – \dfrac{1}{m_1} = -1$$ We already found $y’ = -\dfrac{y}{x}$ in part (a), and so we want to find points on the curve such that

\begin{align*} y’ = -\frac{y}{x} &= -1 \\ \\ y&= x \end{align*} To find the points on the curve when this is true, we substitute this requirement into our original relation:

\begin{align*} y^2x^2 &= 256 \\ \\ (x^2)(x^2) &= 256 \\ \\ x^4 &= 256 \\ x &= \pm 4 \end{align*} We know from our $y’$-requirement above we must have $y = x$, and so

$$y = x = \pm 4$$ We can double check the $y$-value by remembering the solutions $x = \pm 4$ and $y = \pm 4$ must satisfy our original relation: $$y^2x^2 = (\pm 4)^2(\pm 4)^2 = (16)(16) = 256 \checkmark$$

Hence the requested points on the curve are (-4, -4), (-4, 4), (4, -4), and (4, 4). $\quad \cmark$

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Question 6: Second derivative of a circle

Consider the circle $x^2 + y^2 = r^2$, where $r$ is a constant. Find $y^″(x)$.

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We’ll first find $y’$ using implicit differentiation:

\begin{align*} \frac{d}{dx}\left[ x^2 + y^2 \right] &= \frac{d}{dx}\left(r^2 \right) &&\text{[Remember that $r$ is a constant.]} \\ 2x + 2yy’ &= 0 \\ \\ 2yy’ &= 2x \\ \\ y’ &= -\frac{x}{y} \end{align*} To take the second derivative, we can use the quotient rule:

\begin{align*} y^″(x) &= -\frac{\left(\dfrac{d}{dx}x\right)y – x\left(\dfrac{d}{dx}y\right) }{y^2} \\ \\ &= -\frac{y -xy’}{y^2} && \left[\text{Recall from above } y’ = -\dfrac{x}{y}\right] \\ \\ &= -\frac{y – x\left(-\dfrac{x}{y}\right)}{y^2} \\ \\ &= -\frac{y^2 + x^2}{y^3} && \left[\text{Recall that } x^2 + y^2 = r^2 \right]\\ \\ &= -\frac{r^2}{y^3} \quad \cmark \end{align*}

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Question 7: Acutal university exam problem #1

Find the equation of the line tangent to the graph of $x^2y^2 + 2y^3 = 3$ at the point (1,3) on the graph.

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To write the equation of a line, we need to know its slope and a point on the line. We are already given a point, (1,3), and so we only need to find its slope. That slope is equal to the slope of the curve, $y’ = \dfrac{dy}{dx}$, at (1,3). Using implicit differentiation is easiest:

\begin{align*} \frac{d}{dx}\left[x^2y^2 + 2y^3\right] &= \frac{d}{dx}(3) \\ \\ \left(\left(\frac{d}{dx}x^2\right)y^2 + x^2\left(\frac{d}{dx}y^2\right) \right)+ \frac{d}{dx}\left(2y^3\right) &= 0 \\ \\ (2x)y^2 + x^2(2y y’) + 6y^2 y’ &= 0 \\ \\ y’\left(2x^2y + 6y^2 \right) &= -2xy^2 \\ \\ y’ &= \frac{-2xy^2}{2x^2y + 6y^2} \end{align*} At the point (1,3):

\begin{align*} \left. y’ \right|_{(1,3)} &= \frac{-2(1)(3)^2}{2(1)^2(3) + 6(3)^2} \\ \\ &= \frac{-18}{6+54} = \frac{-18}{60} = \frac{-3}{10} \end{align*} Knowing the slope of the tangent line, $m = \left. y’ \right|_{(1,3)} = -\dfrac{3}{10}$, and the point on the line (1,3), we can write its equation in point-slope form:

\begin{align*} y – y_0 &= m(x – x_0) \\ \\ y -3 &= -\dfrac{3}{10}(x-1) \\ \\ y &= -\dfrac{3}{10}x +\dfrac{33}{10} \quad \cmark \end{align*}

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Question 8: Actual university exam problem #2

Consider the curve $x + xy + 2y^2 = 6$.

(a) Find an expression for the slope of the curve at any point $(x, y)$.

(b) Write an equation for the line tangent to the point (2,1).

(c) Find the coordinates of all other points on this curve with slope equal to the slope at (2,1).

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Solution SummarySolution (a) DetailSolution (b) DetailSolution (c) Detail

(a) $y’ = \dfrac{-(1 + y)}{x + 4y}$

(b) $y – 1 = -\dfrac{1}{3}(x-2) $

(c) (6, -3)

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The slope at any point is given by $y’ = \dfrac{dy}{dx}$. We will use implicit differentiation to find $y’$:

\begin{align*} \frac{d}{dx}\left[x + xy + 2y^2\right] &= \frac{d}{dx}(6) \\ \\ \frac{d}{dx}(x) + \left(\left(\frac{d}{dx}x\right)y + x\left(\frac{d}{dx}y\right) \right) + \frac{d}{dx}\left(2y^2\right) &= 0 \\ \\ 1 + \left(y + xy’ \right) + 4yy’ &= 0 \\ \\ y’\left(x + 4y \right) &= -(1 + y) \\ \\ y’ &= \frac{-(1 + y)}{x + 4y} \quad \cmark \end{align*}

The line tangent to the curve at the point (2,1) has slope, using our result from part (a):

\begin{align*} y’ &= \frac{-(1 + y)}{x + 4y} \\ \\ \left. y’ \right|_{(2,1)} &= \frac{-(1 + 1)}{2 + 4(1)} = \frac{-2}{6} = -\frac{1}{3} \\ \\ \end{align*} The line contains the point (2,1) and has slope $m = \left. y’ \right|_{(2,1)} = -\dfrac{1}{3}$, and so we can write it in point-slope form as:

\begin{align*} y – y_0 &= m(x – x_0) \\ \\ y – 1 &= -\frac{1}{3}(x-2) \quad \cmark \end{align*}

We found in part (a) the slope at any point $(x,y$): $$$y’ = \dfrac{-(1 + y)}{x + 4y}$$ We want to find all points on the curve where this slope equals the slope at (2,1), which we found in part (b) to be $\left. y’ \right|_{(2,1)} = -\dfrac{1}{3}$. Let’s call a point that satisfies this requirement $(a,b)$, such that

\begin{align*} y’ = \frac{-(1 + b)}{a + 4b} &= -\frac{1}{3} \\ \\ 3(1+b) &= a + 4b \\ \\ 3 + 3b &= a + 4b \\ \\ 3 – b &= a \end{align*} The point $(a,b)$ must lie on the curve, and so satisfy our original relation $x + xy + 2y^2 = 6$:

\begin{align*} a + ab + 2b^2 &= 6 \\ \\ (3-b) + (3-b)b + 2b^2 &= 6 \\ \\ 3 – b + 3b – b^2 + 2b^2 – 6 &= 0 \\ \\ b^2 + 2b -3 &= 0 \\ \\ (b+3)(b-1) &= 0 \end{align*} Hence the point with $y = -3$ has the same slope as the point with $y = 1$ (which is the point whose slope we’re trying to match).

To find the $x$-value $a$ that corresponds to $y = b = -3$, we can use the relation we found above $3 – b = a$: \begin{align*} 3 – b &= a \\ \\ 3 – (-3) &= a \\ \\ 6 &= a \end{align*} So the point that has the same slope as that of (2,1) is (6, -3). $\quad \cmark$

Just to double check, let’s make sure the point (6, -3) is in fact on the curve:

\begin{align*} x + xy + 2y^2 &= 6 \\ \\ 6 + (6)(-3) + 2(-3)^2 &= 6 – 18 + 18 = 6 \checkmark \end{align*}

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