Implicit Differentiation

The real problem of humanity is the following: We have Paleolithic emotions, medieval institutions and godlike technology. And it is terrifically dangerous, and it is now approaching a point of crisis overall, Edward O. Wilson.

Recall

The derivative of a function at a specific input value describes the instantaneous rate of change. Geometrically, when the derivative exists, it is the slope of the tangent line to the graph of the function at that point. Algebraically, it represents the ratio of the change in the dependent variable to the change in the independent variable as the change approaches zero.

Definition. A function f(x) is differentiable at a point “a” in its domain, if its domain contains an open interval around “a”, and the following limit exists $f'(a) = L = \lim _{h \to 0}{\frac {f(a+h)-f(a)}{h}}$. More formally, for every positive real number ε, there exists a positive real number δ, such that for every h satisfying 0 < |h| < δ, then the following inequality holds |L-$\frac {f(a+h)-f(a)}{h}$|< ε.

Basic important derivatives

1. Power Rule: $\frac{d}{dx}(x^n) = nx^{n-1}$ (works for any real n).
2. Sum Rule: $\frac{d}{dx}(f(x) + g(x)) = \frac{d}{dx}(f(x)) + \frac{d}{dx}(g(x))$
3. Difference Rule: $\frac{d}{dx}(f(x) - g(x)) = f'(x) - g'(x)$
4. Constant Multiple Rule: $\frac{d}{dx}(c \cdot f(x)) = c \cdot f'(x)$ where c is a constant.
5. Product Rule: $\frac{d}{dx}(f(x) \cdot g(x)) = f'(x)g(x) + f(x)g'(x)$.
6. Quotient Rule: $\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}$, provided $g(x) \ne 0$.
7. Chain Rule: $\frac{d}{dx}(f(g(x))) = f'(g(x)) \cdot g'(x)$
8. $\frac{d}{dx}(e^x) = e^x, \frac{d}{dx}(\ln(x)) = \frac{1}{x}, \frac{d}{dx}(\sin(x)) = \cos(x), \frac{d}{dx}(\cos(x)) = -\sin(x), \frac{d}{dx}(\tan(x)) = \sec^2(x), \frac{d}{dx}(\arcsin(x)) = \frac{1}{\sqrt{1 - x^2}}, \frac{d}{dx}(\arccos(x)) = -\frac{1}{\sqrt{1 - x^2}}, \frac{d}{dx}(\arctan(x)) = \frac{1}{1 + x^2}.$

 

Implicit Differentiation

An explicit function expresses y directly in terms of x: y = f(x), e.g., $y = x^2 + 3x - 5, y = \sin(x), y = e^x + \ln(x)$.

In mathematics, some equations involving two variables, x and y, do not explicitly define y as a function of x. This means that even though y can be expressed as a function of x implicitly, the equation might be too complex or even impossible to rearrange in a way that solves for y explicitly in terms of x or give multiple branches. These equations are said to be defined implicitly. An implicit equation defines a relationship between x and y without isolating y: F(x, y) = 0, e.g., - $x^2 + y^2 = 1$ (circle), $x^2 + xy + y^3 = 0, \sin(xy) = x + y, e^y + xy = e$ (cannot be solved for $y$ explicitly).

For example, consider an equation like x² + y² = 1. This equation implicitly defines a relationship between x and y, but it does not express y directly as a function of x (such as y = f(x)). However, there might still exist a function y=f(x) that satisfies this equation for each x.

The technique of implicit differentiation allows us to find the derivative of y with respect to x (denoted as $\frac{dy}{dx}$) without needing to solve the equation explicitly for y.

General Recipe

Given an equation involving x and y, say F(x, y) = 0, to compute $\dfrac{dy}{dx}$:

1. Treat y as a function of x, y = y(x). Differentiate both sides of the defining equation with respect to x: $\frac{d}{dx}F(x,y) = 0.$
2. Apply usual rules (sum, product, chain, etc.). Every time you differentiate y, multiply by $y' = \dfrac{dy}{dx}$.
3. Collect all terms involving y’ on one side.
4. Factor out y'.
5. Solve algebraically for y'.

• Find the slope of the tangent line to the curve x² + y² = 25 at the point (3, -4).

Even though y isn’t isolated, we treat it as y(x) and apply the chain rule. This avoids solving for y explicitly (which would require splitting into $y=\pm\sqrt{25-x^2}$).

Differentiate Both Sides with Respect to x (power and chain rules): 2x +2y·y’ = 0 ⇒[Solve for y’] 2y·y’ = -2x. y·y’ = -x ⇒ y’ = $\frac{-x}{y}$ ⇒ [Evaluate y’ at the point (3, -4)] y’(3) = $\frac{-x}{y} = \frac{-3}{-4} = \frac{3}{4}$. Therefore, the slope of the tangent line at (3, -4) is $\frac{3}{4}$.

The curve $x^2+y^2$ is a circle with radius 5. The radius to the point (3, −4) has a slope of $\frac{-4-0}{3-0}=\frac{-4}{3}$ (the radius connects the center (0, 0) to (3, -4)). Tangent lines are perpendicular to radii (tangent ⟂ radius), so the tangent slope should be the negative reciprocal of the radius slope $\frac{-4}{3}$, which is $\frac{3}{4}$. This matches our result!

• The circle x² + y² = 1 (Implicit definition) relates x and y. It does not give a single-valued function y =f(x) instead, y² = 1 - x² ⇨ $y=\pm\sqrt{1-x^{2}}$ (Explicit definition). This gives two functions or branches (upper and lower semicircles). We must choose one branch.

Yet the curve still has a well-defined slope of the tangent line at most points, and we often want $\frac{dy}{dx}$ even if we don’t have a nice formula y = f(x). That is exactly what implicit differentiation is for.

Let’s take the positive branch, $y=+\sqrt{1-x^{2}}=(1-x^{2})^{\frac{1}{2}}$

Method 1: Explicit Differentiation. y’ = $\frac{1}{2}(1-x^{2})^{\frac{-1}{2}}(-2x) = -x(1-x^{2})^{\frac{-1}{2}}=\frac{-x}{\sqrt{1-x^{2}}}$

Method 2: Implicit Differentiation. Differentiate both side with respect to x, x² + y² = 1: $\frac{d}{dx}(x^{2}+y^{2}=1)$, $2x+2yy'=0$. Solve for y’: $y'=\frac{-x}{y} = \frac{-x}{\pm\sqrt{1-x^2}}.$ The implicit way does not need to take only one branch.

• A More Complicated Implicit Function. Differentiate x² +xy + y³ = 0 implicitly to find $\frac{dy}{dx}$.

1. Start with x² +xy + y³ = 0.
2. Differentiate both sides: $2x +y +x·\frac{dy}{dx} \text{ (product rule) } + 3y^2·\frac{dy}{dx} \text{ (chain rule) }$
3. Collect $\frac{dy}{dx}$ terms: $+x·\frac{dy}{dx} + 3y^2·\frac{dy}{dx} = -2x-y$
4. Factor and solve for $\frac{dy}{dx}$: $\boxed{\frac{dy}{dx} = -\frac{2x+y}{x+3y^2}}$.

• Second Derivative from an Implicit Relation. For the curve x³ + y³ = 9, find $\frac{dy}{dx}, \frac{d^2y}{dx^2}$

1. Start with: x³ + y³ = 9.
2. Differentiate once: $\frac{d}{dx}(x^{3}+y^3=0) \implies 3x^2+3y^2\frac{dy}{dx} = 0 \implies \frac{dy}{dx}=\frac{-3x^2}{3y^2}=\frac{-x^2}{y^2}$
3. Differentiate $y'=\frac{-x^2}{y^2}$ using the quotient rule:

$$ \begin{aligned} \frac{d^2y}{dx^2} &=\frac{-2x·y^2-(-x^2)(2y\frac{dy}{dx})}{(y^2)^2} \\[2pt] &=\frac{-2xy^2+2x^2y\frac{dy}{dx}}{y^4} \\[2pt] &[\text{Substitute First Derivative}] \\[2pt] &=\frac{-2xy^2+2x^2y·\frac{-x^2}{y^2}}{y^4} =\frac{-2xy^2-\frac{2x^4}{y}}{y^4} \\[2pt] &=\frac{\frac{-2xy^3-2x^4}{y}}{y^4}=\frac{-2xy^3-2x^4}{y^5} \\[2pt] &=\frac{-2x(y^3+x^3)}{y^5} \\[2pt] &=[\text{But from the original equation:} x^3+y^3=9]\frac{-2x\cdot 9}{y^5} = \frac{-18x}{y^5} \end{aligned} $$

• A Messy Explicit Solution vs a Simple Implicit One. Differentiate y⁴ + xy² -2 = 0 with respect to x.

Explicit approach (not recommended): Solve for $y^2$, $y^{2}=\frac{-x\pm\sqrt{x^{2}+8}}{2}, y=\pm\sqrt{\frac{-x\pm\sqrt{x^{2}+8}}{2}}$. This is extremely complicated to differentiate!

Implicit approach. Differentiate: 4y³y’+ y² + 2yy’x = 0. Collect terms: (4y³+2xy)y’ + y² = 0. Solve $\boxed{y' = \frac{-y^{2}}{4y^{3}+2xy}=\frac{-y}{4y^2+2x}}$.

• Power Rule for Rational Exponents via Implicit Differentiation. If u(x) = x^a, then u’(x)= ax^a-1, where a=m/n, and m and n are both integers, n > 0, e.g., Square Root, y = $\sqrt{x} = x^{\frac{1}{2}} ⇒ y' = \frac{1}{2}x^{\frac{1}{2}-1} = \frac{1}{2}x^{\frac{-1}{2}} = \frac{1}{2\sqrt{x}}$.

Write in implicit form: $y = x^{\frac{m}{n}} \iff y^n = x^m$. Differentiate both sides with respect to x: $\frac{d}{dx}y^{n} = \frac{d}{dx}x^{m}$.

Right side (ordinary power rule): $\frac{d}{dx}x^{m} = mx^{m-1}$. Left side (chain rule): $\frac{d}{dx}y^{n} = ny^{n-1}\frac{dy}{dx}$

So: $ny^{n-1}\frac{dy}{dx} = mx^{m-1}$. Solve for y’: $\frac{dy}{dx} = \frac{m}{n} \frac{x^{m-1}}{y^{n-1}} =[\text{Substitute } y = x^{\frac{m}{n}}] \frac{m}{n} \frac{x^{m-1}}{(x^\frac{m}{n})^{n-1}} = ax^{m-1-(n-1)\frac{m}{n}}$

Simplify the exponent: $m-1-(n-1)\frac{m}{n} = m -1 -m + \frac{m}{n} = \frac{m}{n} -1 = a -1$

Conclude: $\boxed{\frac{dy}{dx} = ax^{a-1}},~ where~ a=\frac{m}{n}$

• Trigonometric Function: $\sin(xy) = x$. Differentiate: $\cos(xy) \cdot \frac{d}{dx}[xy] = 1 \implies \cos(xy) \cdot \left(y + x\frac{dy}{dx}\right) = 1$. Solve: $y\cos(xy) + x\cos(xy)\frac{dy}{dx} = 1 \implies x\cos(xy)\frac{dy}{dx} = 1 - y\cos(xy)$. Conclude: $\boxed{\frac{dy}{dx} = \frac{1 - y\cos(xy)}{x\cos(xy)}}$

• Exponential and Logarithmic. $e^{xy} = x + y$. Differentiate: $e^{xy} \cdot \left(y + x\frac{dy}{dx}\right) = 1 + \frac{dy}{dx}$. Expand: $ye^{xy} + xe^{xy}\frac{dy}{dx} = 1 + \frac{dy}{dx}$. Collect $\frac{dy}{dx}$ terms: $xe^{xy}\frac{dy}{dx} - \frac{dy}{dx} = 1 - ye^{xy} \implies (xe^{xy} - 1)\frac{dy}{dx} = 1 - ye^{xy}$. Solve for y’: $\boxed{\frac{dy}{dx} = \frac{1 - ye^{xy}}{xe^{xy} - 1}}$

• Equation with y in Multiple Terms: $x^2y^3 - xy = 10$

Differentiate using product rule on each term: For $x^2y^3$: $\frac{d}{dx}[x^2y^3] = 2xy^3 + x^2 \cdot 3y^2\frac{dy}{dx} = 2xy^3 + 3x^2y^2\frac{dy}{dx}$. For $xy$: $\frac{d}{dx}[xy] = y + x\frac{dy}{dx}$

Combine: $2xy^3 + 3x^2y^2\frac{dy}{dx} - y - x\frac{dy}{dx} = 0$. Collect $\frac{dy}{dx}$ terms: $(3x^2y^2 - x)\frac{dy}{dx} = y - 2xy^3$. Solve for y’: $\boxed{\frac{dy}{dx} = \frac{y - 2xy^3}{3x^2y^2 - x} = \frac{y(1 - 2xy^2)}{x(3xy^2 - 1)}}$.

Finding Tangent and Normal Lines

At a specific point $(x_0, y_0)$ on a curve, the slope of the curve is given by the derivative $\frac{dy}{dx}$ evaluated at that point, denoted as $m = \frac{dy}{dx}\big|_{(x_0, y_0)}$.

1. Tangent Line. This line touches the curve at $(x_0, y_0)$ and has the same slope $m$: $y - y_0 = m(x - x_0) = \frac{dy}{dx}\bigg|_{(x_0, y_0)} (x - x_0)$
2. Normal Line. This line is perpendicular to the tangent line at $(x_0, y_0)$. Its slope is the negative reciprocal of the tangent slope ($-\frac{1}{m}$): $y - y_0 = -\frac{1}{m}(x-x_0) = -\frac{1}{\frac{dy}{dx}\big|_{(x_0, y_0)}} (x - x_0)$

Example: Find the equations of the tangent and normal lines to the ellipse $\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1$ at the point $(2\sqrt{2}, \frac{3}{\sqrt{2}})$

1. First, let’s ensure the point $P$ is actually on the ellipse: $\frac{(2\sqrt{2})^2}{16} + \frac{(3/\sqrt{2})^2}{9} = \frac{8}{16} + \frac{9/2}{9} = \frac{1}{2} + \frac{1}{2} = 1$.
2. Implicit differentiation. Differentiate both sides of the equation with respect to x. Remember to apply the Chain Rule to $y$ terms (multiply by $\frac{dy}{dx}$). $\frac{2x}{16} + \frac{2y}{9}\frac{dy}{dx} = 0 \implies[\text{Simplify}] \frac{x}{8} + \frac{2y}{9}\frac{dy}{dx} = 0$. Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = -\frac{x}{8} \cdot \frac{9}{2y} = -\frac{9x}{16y}$
3. Evaluate the slope: Tangent Slope: $m = \frac{dy}{dx}\bigg|_{(2\sqrt{2}, 3/\sqrt{2})} = -\frac{9 \cdot 2\sqrt{2}}{16 \cdot \frac{3}{\sqrt{2}}} = -\frac{18\sqrt{2}}{\frac{48}{\sqrt{2}}} = -\frac{18\sqrt{2} \cdot \sqrt{2}}{48} = -\frac{36}{48} = -\frac{3}{4}$. Normal Slope: $m_{\perp} = -\frac{1}{(-3/4)} = \frac{4}{3}$
4. Write the Equations. Now, use the point-slope form $y - y_0 = m(x - x_0)$ with our specific slopes. Tangent Line: $y - \dfrac{3}{\sqrt{2}} = -\dfrac{3}{4}\left(x - 2\sqrt{2}\right)$. Normal Line: $y - \dfrac{3}{\sqrt{2}} = \dfrac{4}{3}\left(x - 2\sqrt{2}\right)$

Logarithmic Differentiation

A special application of implicit differentiation for products, quotients, or variable exponents. For $y = f(x)$, e.g., $y = x^x, \quad x > 0$:

1. Take natural logarithms: $\ln(y) = \ln(f(x))$, $\ln(y) = \ln(x^x)$.
2. Simplify using logarithms properties, $\ln(y) = x\ln(x)$
3. Differentiate implicitly, $\frac{1}{y}\frac{dy}{dx} = \ln(x) + x \cdot \frac{1}{x} = \ln(x) + 1$
4. Solve for y’, $\frac{dy}{dx} = y(\ln(x) + 1)$
5. Substitute back $y = f(x)$, $\frac{dy}{dx} = x^x(\ln(x) + 1)$

Another example: $y = \frac{x^2\sqrt{x+1}}{(x-2)^3}$

1. Take logarithm: $\ln(y) = 2\ln(x) + \frac{1}{2}\ln(x+1) - 3\ln(x-2)$
2. Differentiate: $\frac{y'}{y} = \frac{2}{x} + \frac{1}{2(x+1)} - \frac{3}{x-2}$
3. Solve: $y' = y\left(\frac{2}{x} + \frac{1}{2(x+1)} - \frac{3}{x-2}\right) = \frac{x^2\sqrt{x+1}}{(x-2)^3}\left(\frac{2}{x} + \frac{1}{2(x+1)} - \frac{3}{x-2}\right)$

Inverse Functions

For inverse functions that are difficult or even impossible to differentiate directly, we can:

1. Rewrite the inverse function as an implicit equation.
2. Apply implicit differentiation.
3. Use trigonometric identities to express the result in terms of $x$.

General strategy:

1. For $y = f^{-1}(x)$, rewrite it as $f(y) = x$.
2. Differentiate both sides with respect to $x$.
3. Apply chain rule: $f'(y) \cdot \dfrac{dy}{dx} = 1$.
4. Solve: $\dfrac{dy}{dx} = \dfrac{1}{f'(y)}$.
5. Express $f'(y)$ in terms of $x$ using identities.

Derivative of Inverse Sine: $\arcsin(x)$

Let $y = \arcsin(x)$, which means $\sin(y) = x$ where $y \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$

1. Differentiate both sides with respect to $x$: $\frac{d}{dx}[\sin(y)] = \frac{d}{dx}[x]$
2. Apply chain rule on the left: $\cos(y) \cdot \frac{dy}{dx} = 1$
3. Solve for $\dfrac{dy}{dx}$: $\frac{dy}{dx} = \frac{1}{\cos(y)}$
4. Express $\cos(y)$ in terms of $x$. Using $\sin^2(y) + \cos^2(y) = 1$: $\cos^2(y) = 1 - \sin^2(y) = 1 - x^2.$ Then, $\cos(y) = \pm\sqrt{1 - x^2}$
5. Determine the sign. Since $y \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$, we have $\cos(y) \geq 0$. Therefore: $\cos(y) = +\sqrt{1 - x^2}$
6. In conclusion, $\boxed{\frac{d}{dx}[\arcsin(x)] = \frac{1}{\sqrt{1 - x^2}}, \quad |x| < 1}$

Derivative of Inverse Tangent: $\arctan(x)$

Let $y = \arctan(x)$, which means $\tan(y) = x$ where $y \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$

1. Differentiate implicitly: $\frac{d}{dx}[\tan(y)] = \frac{d}{dx}[x]$
2. Apply chain rule on the left: $\sec^2(y) \cdot \frac{dy}{dx} = 1$
3. Solve for $\dfrac{dy}{dx}$: $\frac{dy}{dx} = \frac{1}{\sec^2(y)} = \cos^2(y)$
4. Express $\cos^2(y)$ in terms of $x$. Using the identity $1 + \tan^2(y) = \sec^2(y) \implies \sec^2(y) = 1 + \tan^2(y) = 1 + x^2 \implies \cos^2(y) = \frac{1}{\sec^2(y)} = \frac{1}{1 + x^2}$
5. In conclusion, $\boxed{\frac{d}{dx}[\arctan(x)] = \frac{1}{1 + x^2}, \quad x \in \mathbb{R}}$

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