Differentiation Techniques: Implicit and Inverse Functions

Implicit Functions and Implicit Differentiation

Calculus deals with functions and their rates of change. Often, functions are given in a form where the dependent variable is explicitly defined in terms of the independent variable. However, relationships between variables are not always expressible in such a clear-cut manner. In such cases, we deal with implicit relations, and a technique called implicit differentiation allows us to find derivatives.

Explicit vs. Implicit Functions

• An explicit function is one where the dependent variable (most commonly $y$) is isolated on one side of the equation, directly expressed as a formula involving only the independent variable (most commonly $x$).

  Example: $y = x^2 + \sin x$, $y = \frac{e^x}{x+1}$, $f(x) = \sqrt{x} + 3x$. In these cases, $y$ or $f(x)$ is clearly a function of $x$ alone.

• An implicit function (or more precisely, an implicit relation that may define one or more functions) is one where the relationship between $x$ and $y$ is given by an equation where $y$ is not explicitly solved for in terms of $x$.

  Example: $x^2 + y^2 = 25$, $x^3 + y^3 = 6xy$, $\sin(xy) = x + y$. In these equations, $x$ and $y$ are mixed together, and $y$ is not directly written as $y = \text{some expression of } x$.

While some implicit relations can be rearranged to solve for $y$ explicitly (e.g., from $x^2+y^2=25$, we get $y^2 = 25-x^2$, so $y = \pm \sqrt{25-x^2}$, which actually defines two separate explicit functions), many cannot be easily or even possibly solved for $y$ algebraically (e.g., $x^3 + y^3 = 6xy$). Implicit differentiation provides a method to find the derivative $\frac{dy}{dx}$ even when we cannot solve for $y$ explicitly.

Implicit Differentiation Process

The technique of implicit differentiation is based on the Chain Rule. When we differentiate an equation involving both $x$ and $y$ with respect to $x$, we treat $y$ as a differentiable function of $x$, say $y=f(x)$. Any term involving $y$ must be differentiated using the Chain Rule, where the "inner" function is $y(x)$ and the "outer" function is whatever operation is applied to $y$. The derivative of $y$ with respect to $x$ is simply denoted by $\frac{dy}{dx}$.

To find $\frac{dy}{dx}$ for an equation implicitly defining $y$ as a function of $x$:

1. Differentiate both sides of the equation with respect to $x$. Treat the equals sign as a balance scale; whatever operation you perform on one side, you must perform on the other. We apply the derivative operator $\frac{d}{dx}$ to every term on both sides of the equation.
2. Apply standard differentiation rules for terms involving only $x$. For terms containing only the variable $x$ (and constants), use the standard rules like the Power Rule, Product Rule, Quotient Rule, etc., just as you would for explicit functions. For instance, $\frac{d}{dx}(x^3) = 3x^2$.
3. Apply the Chain Rule for terms involving $y$. This is the key step in implicit differentiation. When differentiating a term that contains $y$, remember that $y$ is a function of $x$. Use the Chain Rule $ \frac{d}{dx}[F(y)] = F'(y) \cdot \frac{dy}{dx} $, where $F$ is the outer function and $y$ is the inner function. Common examples:
   • $\frac{d}{dx}(y^n) = n y^{n-1} \cdot \frac{dy}{dx}$ (Using General Power Rule where $g(x)=y$ and $g'(x)=\frac{dy}{dx}$)
   • $\frac{d}{dx}(\sin y) = \cos y \cdot \frac{dy}{dx}$ (Using Chain Rule where $F(u)=\sin u$ and $u=y$)
   • $\frac{d}{dx}(e^y) = e^y \cdot \frac{dy}{dx}$ (Using Chain Rule where $F(u)=e^u$ and $u=y$)
   • When a term involves both $x$ and $y$ multiplied, like $x \cdot y$, use the Product Rule: $\frac{d}{dx}(x \cdot y) = x \cdot \frac{d}{dx}(y) + y \cdot \frac{d}{dx}(x) = x \frac{dy}{dx} + y(1) = x \frac{dy}{dx} + y$.
4. Algebraically solve for $\frac{dy}{dx}$. The resulting equation after differentiation will usually contain several terms, some with $\frac{dy}{dx}$ and some without. Treat $\frac{dy}{dx}$ as an algebraic variable and rearrange the equation to isolate it. Collect all terms containing $\frac{dy}{dx}$ on one side of the equation and all other terms on the other side. Factor out $\frac{dy}{dx}$ from the terms on one side, and then divide by the remaining factor to solve for $\frac{dy}{dx}$.

The final expression for $\frac{dy}{dx}$ obtained through implicit differentiation will often be in terms of both $x$ and $y$, which is perfectly acceptable and is necessary because the derivative's value depends on the specific point $(x,y)$ on the curve.

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Derivatives of Inverse Functions (General Principle)

Calculus provides tools to analyze the rate of change of functions. When we have a function $f(x)$ that is one-to-one (meaning it has a unique inverse function), we can also find the derivative of its inverse function. There is a powerful theorem that relates the derivative of the inverse function to the derivative of the original function.

Inverse Functions

Two functions, let's call them $f$ and $g$, are said to be inverse functions if applying one function reverses the action of the other. Formally, $g$ is the inverse of $f$ if:

1. $f(g(x)) = x$ for every $x$ in the domain of $g$, and
2. $g(f(y)) = y$ for every $y$ in the domain of $f$.

If such a function $g$ exists, it is unique and is denoted by $f^{-1}$. So, if $g$ is the inverse of $f$, we write $g = f^{-1}$.

For a function $f$ to have an inverse $f^{-1}$, it must be one-to-one (also called injective). A function is one-to-one if $f(x_1) = f(x_2)$ implies $x_1 = x_2$. Graphically, a function is one-to-one if no horizontal line intersects its graph more than once (the Horizontal Line Test).

If $(a, b)$ is a point on the graph of $y = f(x)$, then $(b, a)$ is a point on the graph of the inverse function $y = f^{-1}(x)$. The graph of $y = f^{-1}(x)$ is the reflection of the graph of $y = f(x)$ across the line $y = x$.

Theorem for the Derivative of an Inverse Function

Suppose $f$ is a differentiable function on an interval $I$. If $f$ has an inverse function $g = f^{-1}$, and if $f'(g(a))$ exists and is not equal to zero, then the inverse function $g$ is differentiable at the point $a$ (which is in the range of $f$), and its derivative is given by:

$g'(a) = \frac{1}{f'(g(a))}$

Using Leibniz notation, if $y = f(x)$ is a differentiable function of $x$, and $x = g(y) = f^{-1}(y)$ is the inverse function differentiable with respect to $y$, then their derivatives are related by:

$\frac{dx}{dy} = \frac{1}{dy/dx}$

This means the rate of change of $x$ with respect to $y$ is the reciprocal of the rate of change of $y$ with respect to $x$, provided $\frac{dy}{dx} \neq 0$. This is evaluated at the corresponding points: $\frac{dx}{dy}$ at $y=a$ equals $\frac{1}{dy/dx}$ at $x=g(a)$.

Derivation using the Chain Rule

We can derive the formula for the derivative of an inverse function using the definition of inverse functions and the Chain Rule.

Let $y = f(x)$ be a differentiable function with an inverse function $g(x) = f^{-1}(x)$. By the definition of an inverse function, applying $f$ to $g(x)$ (or $f^{-1}$) gives the original input $x$:

Now, differentiate both sides of this equation with respect to $x$. On the left side, we have a composite function $f(g(x))$, so we must use the Chain Rule:

Apply the Chain Rule on the left side ($f'(g(x)) \cdot g'(x)$) and differentiate $x$ on the right side ($\frac{d}{dx}(x) = 1$):

Now, we want to find $g'(x)$, so we solve for $g'(x)$ by dividing both sides by $f'(g(x))$, provided that $f'(g(x)) \neq 0$:

This formula gives the derivative of the inverse function $g'(x)$ for any $x$ in its domain. Evaluating this formula at a specific point $x=a$ gives the theorem:

$g'(a) = \frac{1}{f'(g(a))}$.

The condition $f'(g(a)) \neq 0$ is important. If $f'(g(a)) = 0$, it means the original function $f$ has a horizontal tangent at the point $(g(a), f(g(a)))$. Since the inverse graph is reflected across $y=x$, the inverse function $g$ will have a vertical tangent at the corresponding point $(f(g(a)), g(a)) = (a, g(a))$, and thus the derivative $g'(a)$ will be undefined (infinite).

Geometric Interpretation

The formula $g'(b) = \frac{1}{f'(a)}$ (where $b=f(a)$ and $a=g(b)$) has a clear geometric meaning. If $(a, b)$ is a point on the graph of $y=f(x)$, the slope of the tangent line at this point is $f'(a)$. The corresponding point on the graph of the inverse function $y=g(x)$ is $(b, a)$, and the slope of the tangent line at this point is $g'(b)$.

Since the graph of $y=g(x)$ is obtained by reflecting the graph of $y=f(x)$ across the line $y=x$, the slopes of the tangent lines at corresponding points are reciprocals of each other.

For example, if the tangent to $f(x)$ at $(a,b)$ has a slope of $m = f'(a)$, the tangent to $g(x)$ at $(b,a)$ will have a slope of $1/m = g'(b)$. This is consistent with the reflection across the line $y=x$, which swaps the roles of horizontal and vertical change.

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Derivatives of Inverse Trigonometric Functions

Inverse trigonometric functions (also known as arc functions) are the inverse functions of the restricted trigonometric functions. For example, $\arcsin x$ is the inverse of $\sin x$ restricted to the interval $[-\pi/2, \pi/2]$. Their derivatives are important standard formulas in calculus and can be derived using the general theorem for the derivative of an inverse function or, more commonly, using implicit differentiation.

Derivation of $\frac{d}{dx}(\arcsin x)$

Let $y = \arcsin x$. By the definition of the inverse sine function, this means $x = \sin y$. The domain of $\arcsin x$ is $[-1, 1]$, and its range is $[-\pi/2, \pi/2]$. For $y$ to be in this range, $\cos y$ is non-negative (it's positive on $(-\pi/2, \pi/2)$ and 0 at the endpoints). We want to find $\frac{dy}{dx}$.

1. Start with the inverse relation: $x = \sin y$.
2. Differentiate both sides of the equation with respect to $x$:

   $\frac{d}{dx}(x) = \frac{d}{dx}(\sin y)$

   [Differentiate both sides w.r.t. $x$]

3. Differentiate the left side: $\frac{d}{dx}(x) = 1$.
4. Differentiate the right side using the Chain Rule (since $y$ is a function of $x$). The outer function is $\sin u$, the inner is $y$. $\frac{d}{du}(\sin u) = \cos u$. Substitute $u=y$ and multiply by $\frac{dy}{dx}$: $\frac{d}{dx}(\sin y) = \cos y \cdot \frac{dy}{dx}$.
5. Substitute these derivatives back into the differentiated equation:

   $1 = \cos y \cdot \frac{dy}{dx}$

   [Substituting derivatives]

6. Solve for $\frac{dy}{dx}$:

   $\frac{dy}{dx} = \frac{1}{\cos y}$

   [Divide by $\cos y$, assuming $\cos y \neq 0$]

7. We need to express $\cos y$ in terms of $x$. We know the identity $\sin^2 y + \cos^2 y = 1$.

   $\cos^2 y = 1 - \sin^2 y$

   [Using identity]

   Since $x = \sin y$, we have $\sin^2 y = x^2$.

   $\cos^2 y = 1 - x^2$

   [Substitute $\sin y = x$]

   Taking the square root:

   $\cos y = \pm \sqrt{1 - x^2}$

8. Since the range of $y = \arcsin x$ is $[-\pi/2, \pi/2]$, the value of $\cos y$ for $y$ in this interval is non-negative (except at the endpoints where $\cos y=0$). Therefore, we take the positive square root: $\cos y = \sqrt{1 - x^2}$ (for $-1 < x < 1$).
9. Substitute this back into the expression for $\frac{dy}{dx}$:

   $\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}$

   [Substituting $\cos y$]

Thus, the derivative of $\arcsin x$ is:

$\frac{d}{dx}(\arcsin x) = \frac{1}{\sqrt{1 - x^2}}$ (for $-1 < x < 1$)

The derivative is undefined at $x=\pm 1$, which correspond to the vertical tangents at the endpoints of the $\arcsin x$ graph.

Derivation of $\frac{d}{dx}(\arctan x)$

Let $y = \arctan x$. By definition, $x = \tan y$. The domain of $\arctan x$ is $(-\infty, \infty)$, and its range is $(-\pi/2, \pi/2)$. We want to find $\frac{dy}{dx}$.

1. Start with the inverse relation: $x = \tan y$.
2. Differentiate both sides with respect to $x$:

   $\frac{d}{dx}(x) = \frac{d}{dx}(\tan y)$

   [Differentiate both sides w.r.t. $x$]

3. Differentiate the left side: $\frac{d}{dx}(x) = 1$.
4. Differentiate the right side using the Chain Rule. The outer function is $\tan u$, the inner is $y$. $\frac{d}{du}(\tan u) = \sec^2 u$. Substitute $u=y$ and multiply by $\frac{dy}{dx}$: $\frac{d}{dx}(\tan y) = \sec^2 y \cdot \frac{dy}{dx}$.
5. Substitute these derivatives back into the differentiated equation:

   $1 = \sec^2 y \cdot \frac{dy}{dx}$

   [Substituting derivatives]

6. Solve for $\frac{dy}{dx}$:

   $\frac{dy}{dx} = \frac{1}{\sec^2 y}$

   [Divide by $\sec^2 y$, assuming $\sec^2 y \neq 0$]

7. We need to express $\sec^2 y$ in terms of $x$. We know the identity $\sec^2 y = 1 + \tan^2 y$.

   $\sec^2 y = 1 + (\tan y)^2$

   [Using identity]

   Since $x = \tan y$, we have $(\tan y)^2 = x^2$.

   $\sec^2 y = 1 + x^2$

   [Substitute $\tan y = x$]

8. Substitute this back into the expression for $\frac{dy}{dx}$:

   $\frac{dy}{dx} = \frac{1}{1 + x^2}$

   [Substituting $\sec^2 y$]

Thus, the derivative of $\arctan x$ is:

$\frac{d}{dx}(\arctan x) = \frac{1}{1 + x^2}$ (for all real $x$)

Summary of Derivatives of Inverse Trigonometric Functions

Here is a consolidated list of the derivatives of the six inverse trigonometric functions:

Notice the pattern: the derivatives of the "co-functions" ($\arccos$, $\text{arccot}$, $\text{arccosec}$) are the negatives of the derivatives of their counterparts ($\arcsin$, $\arctan$, $\text{arcsec}$).

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