After learning the rules (power rule, product and quotient rules, chain rule, etc.) and doing numerous exercises, a student realizes that the mechanics of Differential Calculus are not that difficult. Actually, they are almost boring, if it wasn’t for that one rule learned after all the other rules: implicit differentiation.
Differentiation is the rate of change of one variable with respect to another. For example, in the simple function \(\boldsymbol{y = 5x}\), the variable \(\boldsymbol{y}\) changes 5 times as fast as the variable \(\boldsymbol{x}\). We say that the rate of change of \(\boldsymbol{y}\) with respect to \(\boldsymbol{x}\) is 5. We indicate this as: \(\boldsymbol{\frac{dy}{dx} = 5}\), or another common notation: \(\boldsymbol{y’ = 5}\).
Functions of the type where only \(\boldsymbol{y}\) is on one side of the \(\boldsymbol{=}\) sign and \(\boldsymbol{x}\) on the other, are called explicit functions. How do we find the rate of change of something like the equation of a circle: \(\boldsymbol{x^{2} + y^{2} = 25}\)? We can solve for \(\boldsymbol{y}\), like this: \(\boldsymbol{y = \mp \sqrt{25\, – x^{2}}}\), and then differentiate the upper half of the circle (\(\boldsymbol{\frac{dy}{dx} = -\frac{x}{\sqrt{25\, -\, x^{2}}}}\)) and the lower half of the circle (\(\boldsymbol{\frac{dy}{dx} = \frac{x}{\sqrt{25\, -\, x^{2}}}}\)) separately.
Example 1
But we don’t necessarily have to proceed that way. We can use Implicit Differentiation. The equation of a circle we saw above, it is implicitly, a function of \(\boldsymbol{x}\).
\begin{align*}
\text{Let’s try it: } \qquad \qquad x^{2} + y^{2} &= 25 \\
\text{Differentiate both sides with respect to } \boldsymbol{x} \text{: } \frac{d}{dx} \left(x^{2}\right) + \frac{d}{dx} \left(y^{2}\right) &= \frac{d}{dx} \left(25\right) \\
2x + 2y \frac{dy}{dx} &= 0 \\
\text{Solving for } \frac{dy}{dx} \text{ we obtain: } \qquad \qquad \frac{dy}{dx} &= -\frac{2x}{2y} = -\frac{x}{y}
\end{align*}
The obvious question is, why does this result \(\mathbf{\left(\frac{dy}{dx} = -\frac{x}{y}\right)}\) is different from the previous one \(\mathbf{\left(\frac{dy}{dx} = \mp \frac{x}{\sqrt{25-x^{2}}}\right)}\)?. After all, they are both derivatives of the same equation of a circle. The equations look different but they provide the same information.
Let’s take a look at the following graph of a circle.
\begin{align*}
\text{At the point } \mathbf{\left(4,3\right)} \text{ we obtain the following slope:} \\
\text{Using the upper half of the explicit form of the derivative: } -\frac{x}{\sqrt{25\, -\, x^{2}}}\Large|\small_{x=4} \normalsize &= -\frac{4}{\sqrt{25 -\, 4^{2}}} = -\frac{4}{3} \\
\text{Using the implicit (simpler) form of the derivative: } \qquad \; -\frac{x}{y}\Large|\small_{x=4, y=3} \normalsize &= -\frac{4}{3}
\end{align*}
\begin{align*}
\text{At the point } \mathbf{\left(4,-3\right)} \text{ we obtain the following slope:} \\
\text{Using the lower half of the explicit form of the derivative: } \frac{x}{\sqrt{25\, -\, x^{2}}}\Large|\small_{x=4} \normalsize &= \frac{4}{\sqrt{25 -\, 4^{2}}} = \frac{4}{3} \\
\text{Using the implicit (simpler) form of the derivative: } \quad -\frac{x}{y}\Large|\small_{x=4, y=-3} \normalsize &= \frac{4}{3}
\end{align*}
And similarly for the any other point on the circle. As you can see, the implicit form of the derivative is simpler to obtain. You may have noticed that it has a drawback and it is this: the derivative is both in \(\boldsymbol{x}\) and \(\boldsymbol{y}\). This means you cannot obtain the slope of an equation if you only know \(\boldsymbol{x}\).
Example 2
Let’s take a look at another example of Implicit Differentiation: Find equations of both tangent lines to the graph of the ellipse \(\normalsize\mathbf{\frac{x^{2}}{4} + \frac{y^{2}}{9} = 1}\) that pass through the point \(\mathbf{\left(4,0\right)}\) not on the graph.
The graph above shows the \(\mathbf{\color{blue}{ ellipse }}\) and the point \(\mathbf{\color{red}{ \left(4,0\right)}}\). The task is to obtain the equation of the two lines that are tangent to the ellipse and pass through that point.
We’ll start by taking the derivative of the ellipse. This will allow us to build an equation for the slope of the tangent lines. Like Prof. Bruce Roberts says in one of The Great Courses videos: “When in doubt, take the derivative”!
\begin{align*}
\frac{d}{dx} \left(\frac{x^{2}}{4} + \frac{y^{2}}{9}\right) &= \frac{d}{dx} \left(1\right) \\
\frac{2x}{4} + \frac{2yy’}{9} &= 0 \\
\frac{yy’}{9} &= \frac{-x}{4} \\
\text{This is the slope equation for any point in the ellipse: } \qquad y’ &= \frac{-9x}{4y} \tag{Slope equation} \\
\text{Combined with the } \frac{rise}{run} \text{ slope equation: } \quad \frac{-9x}{4y} &= \frac{y\, – 0}{x\, – 4} \\
\text{We cross-multiply and obtain: } -9x \cdot \left(x\, – 4\right) = 4y^{2} \boldsymbol{\Rightarrow} 4y^{2} &= 36x\, – 9x^{2} \\
\text{From the original ellipse equation: } \frac{x^{2}}{4} + \frac{y^{2}}{9} = 1 \boldsymbol{\Rightarrow} 9x^{2} + 4y^{2} = 36 \boldsymbol{\Rightarrow} 4y^{2} &= 36\, – 9x^{2} \\
\text{We see from these last two lines that: } \quad 36x &= 36 \boldsymbol{\Rightarrow} \mathbf{x = 1} \tag{x-coordinate}
\end{align*}
If \(\mathbf{x = 1}\), then we solve for \(\mathbf{y}\) in the equation for the ellipse:
\begin{align*}
\frac{\left(1\right)^{2}}{4} + \frac{y^{2}}{9} = 1 \boldsymbol{\Rightarrow} \frac{y^{2}}{9} &= \frac{3}{4} \boldsymbol{\Rightarrow} y = \pm \sqrt{\frac{27}{4}} \boldsymbol{\Rightarrow} \mathbf{y = \pm \frac{3\sqrt{3}}{2}} \tag{y-coordinate} \\
\text{The points on the ellipse are: } \qquad \mathbf{\left(1, \pm \frac{3\sqrt{3}}{2}\right)}
\end{align*}
The next task is to get the specific slopes of the tangent lines. But for this, we already have an equation:
\begin{align*}
y’ &= \frac{-9x}{4y} \\
&= \frac{-9\left(1\right)}{4 \cdot \left(\pm \frac{3\sqrt{3}}{2}\right)} = \frac{\mp 9}{6\sqrt{3}} = \mp \frac{\sqrt{3}}{2}
\end{align*}
The final task is to build the tangent lines:
\begin{align*}
y\, – \frac{3\sqrt{3}}{2} = -\frac{\sqrt{3}}{2} \cdot \left(x\, – 1\right) \; \mathbf{\Rightarrow} \; y &= -\frac{\sqrt{3}}{2} x + 2\sqrt{3} \\
\text{And similarly: } \qquad y &= \frac{\sqrt{3}}{2} x + 2\sqrt{3}
\end{align*}
The graph above shows the lines tangent to the ellipse that pass through the point \(\mathbf{\left(4,0\right)}\).
Example 3
The last example is the most interesting because here we have both \(\boldsymbol{x}\) and \(\boldsymbol{y}\) as exponents. The task is to find an equation of the tangent line at the given point: \(\normalsize 1 + \ln \left(x \cdot y\right) = e^{x\, – y} \text{, at the point } \left(1,1\right)\).
Let’s start by looking at a graph of the equation and the point \(\mathbf{\color{red}{ \left(1,1\right)}}\)
\begin{align*}
\text{Move everything to one side of the equal sign: } 1 + \ln \left(x \cdot y\right)\, – e^{x\, -\, y} &= 0 \\
\text{Differentiate the equation: } \frac{1}{x \cdot y} \cdot \left(x \cdot y’ + y\right)\, -\, e^{x\, -\, y} \cdot \left(1\, -\, y’\right) &= 0 \\
\text{And now, solve for } \mathbf{y’} \text{: } \frac{y’}{y} + \frac{1}{x}\, -\, e^{x\, -\, y} + e^{x\, -\, y} \cdot y’ &= 0 \\
y’ \cdot \left(\frac{1}{y} + e^{x\, -\, y}\right) &= e^{x\, -\, y}\, -\, \frac{1}{x} \\
y’ = \frac{e^{x\, -\, y}\, -\, \frac{1}{x}}{\frac{1}{y} + e^{x\, -\, y}} = \frac{\frac{x \cdot e^{x\, -\, y}\, -\, 1}{x}}{\frac{1 + y \cdot e^{x\, -\, y}}{y}} &= \frac{y \cdot \left(x \cdot e^{x\, -\, y}\, -\, 1\right)}{x \cdot \left(y \cdot e^{x\, -\, y} + 1\right)}
\end{align*}
Now, to build an equation of the tangent line at the point \(\mathbf{\left(1,1\right)}\), first we calculate the slope: \(\normalsize \frac{1 \cdot \left(1 \cdot e^{1\, -\, 1}\, -\, 1\right)}{1 \cdot \left(1 \cdot e^{1\, -\, 1} + 1\right)} = \frac{0}{2} = 0\).
\[\normalsize y\, -\, 1 = 0 \cdot \left(x\, -\, 1\right) \Rightarrow y = 1\].
And here is a final graph with the tangent line at \(\mathbf{y = 1}\).
Quiz #7
Because of the extent of the post and the answers to Quiz #6, I am going to beg off this week. My apologies. I hope to have a new Quiz next time.
Answer to Quiz #6
Last time we asked what is the domain of the following functions, their concavity, and locate the x- and y-coordinates of their minimum and/or maximum:
\[y = x \ln{x} \quad \text{ and } \quad y =\frac{\ln{x}}{x}\]
Part One
Let’s start with \(\mathbf{y = x \ln{x}}\):
Domain: because logarithms are valid for values \(\mathbf{\gt 0}\), the answer is: \(\mathbf{\left(0,\infty\right)}\).
\begin{align*}
\text{The next step is to take the first derivative: } \frac{dy}{dx} = x \cdot \frac{1}{x} + \ln{x} &= 1 + \ln{x} \\
\text{and solve for zero: } 1 + \ln{x} = 0 \text{ when } x &= \frac{1}{e} \\
x \ln{x}\Large|\small_{x=\frac{1}{e}} &= -\frac{1}{e} \\
\text{Now we know that } \left(\frac{1}{e},-\frac{1}{e}\right) \text{ is either a minimum or a maximum.} \\
\text{Let’s take the second derivative: } \qquad \qquad \frac{d^{2}y}{dx^{2}} &= \frac{1}{x} \\
\text{Sign of the 2nd derivative at } x = \frac{1}{e} \text{ is: } \gt 0
\end{align*}
Now, we have all the answers. \(\mathbf{y = x \ln{x}}\) is concave up and \(\mathbf{\left(\frac{1}{e},-\frac{1}{e}\right)}\) is a minimum. Here’s the graph of the function.
Part Two
Let’s answer the same questions about the second equation: \(\mathbf{y = \frac{\ln{x}}{x}}\).
Domain: because logarithms are valid for values \(\mathbf{\gt 0}\), the answer is: \(\mathbf{\left(0,\infty\right)}\).
\begin{align*}
\text{First derivative: } \frac{dy}{dx} &= \frac{d}{dx}\left(\frac{\ln{x}}{x}\right) \\
&= \frac{x \cdot \left(\frac{1}{x}\right)\, -\, \ln{x}}{x^{2}} = \frac{1\, -\, \ln{x}}{x^{2}} \\
\text{Solve for zero: } \frac{1\, -\, \ln{x}}{x^{2}} &= 0 \text{, } \quad \mathbf{x = e} \\
\frac{\ln{x}}{x}\Large|\small_{x=e} &= \frac{1}{e} \\
\text{Now we know that } \left(e,\frac{1}{e}\right) \text{ is either a minimum or a maximum.} \\
\text{Let’s take the second derivative: } \frac{d^{2}y}{dx^{2}} &= \frac{x^{2} \cdot \left(-\frac{1}{x}\right) – \left(1\, -\, \ln{x}\right) \cdot \left(2x\right)}{x^{4}} = \frac{2x \ln{x}\, -\, 3x}{x^{4}} = \frac{2 \ln{x}\, -\, 3}{x^{3}} \\
\text{Sign of the 2nd derivative at } x = e \text{ is: } \lt 0
\end{align*}
Just like before, we have all the answers. \(\mathbf{y = \frac{\ln{x}}{x}}\) is concave down and \(\mathbf{\left(e,\frac{1}{e}\right)}\) is a maximum. Here’s the graph of the function.
If you find any errors in this post, or have some thoughts about it, please, email me at [email protected] .