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Section 3.7 Implicit Differentiation

Recall differentiating \(y=(x^2+1)^7\text{;}\) we've seen how \(\frac{dy}{dx} = 7(x^2+1)^6 \cdot 2x\text{.}\)

More explicitly, let \(u=x^2+1\text{,}\) so that \(\frac{du}{dx} = 2x\text{.}\) Now we write

\begin{equation*} y=u^7 \end{equation*}

and then

\begin{equation*} \frac{dy}{dx} = 7u^6 \frac{du}{dx} = 7 (x^2+1)^6 2x \end{equation*}

A way you can think of it:

\begin{equation*} \frac{d}{dx} f(u) = f'(u) \frac{du}{dx} \end{equation*}
\begin{equation*} \frac{d}{dx} f(x) = f'(x) \frac{dx}{dx} = f'(x) \end{equation*}

Example 3.7.1.

Consider the circle \(x^2+y^2=9\text{.}\) Note that this is not a function, but we consider \(y\) to be an implicit function of \(x\text{.}\) Find the slope of the circle at \((\sqrt{3},-\sqrt{6})\text{.}\)

Answer.

\(\frac{dy}{dx}= -\frac{x}{y}\text{;}\) \(m=\frac{1}{\sqrt{2}}\)

Example 3.7.2.

  1. Find \(\frac{dy}{dx}\text{:}\) \(5x^2y+6xy^3 = 8\text{.}\)

  2. Find \(\frac{dy}{dx}\text{:}\) \(x-\tan(xy) = 2\text{.}\)

  3. Find \(\frac{ds}{dt}\text{:}\) \((3st-4)^3=10s\text{.}\)

  4. Find \(\frac{dy}{dx}\text{:}\) \(e^{7x}=\sin(x+5y)\text{.}\)

Answer.

  1. \(\displaystyle \displaystyle \frac{dy}{dx} = - \frac{10xy+6y^3}{5x^2+18xy^2}\)

  2. \(\displaystyle \displaystyle \frac{dy}{dx} = \frac{1-y\sec^2(xy)}{x\sec^2(xy)}\)

  3. \(\displaystyle \displaystyle \frac{ds}{dt} = \frac{9s(3st-4)^2}{10-9t(3st-4)^2}\)

  4. \(\displaystyle \displaystyle \frac{dy}{dx} = \frac{7e^{7x} - \cos(x+5y)}{5\cos(x+5y)}\)

Example 3.7.3.

Find \(\frac{dy}{dx}\) and \(\frac{d^2y}{dx^2}\text{:}\)

\begin{equation*} x^{4/5}+y^{4/5}=89 \end{equation*}
Answer.

\(\displaystyle \frac{dy}{dx} = \frac{-y^{1/5}}{x^{1/5}}\text{;}\) \(\displaystyle \frac{d^2y}{dx^2} = \frac{y^{1/5}}{5x^{2/5}}\left(y^{-4/5}+x^{-4/5}\right)\)

Definition 3.7.4. Normal.

A line is normal to a curve at a point if it passes through the point and the slope of the line is the negative reciprocal of the slope of the curve at that point.

Example 3.7.5.

Find the equations of the lines that are tangent and normal to the devil's curve:

\begin{equation*} y^4-4y^2=x^4-9x^2 \end{equation*}

at \((-3,2)\text{.}\)

Answer.

\(\displaystyle\frac{dy}{dx} = \frac{4x^3-18x}{4y^3-8y}\)

tangent: \(y=-\frac{27}{8}x-\frac{65}{8}\text{;}\) normal: \(y=\frac{8}{27}x+\frac{26}{9}\)