Derivatives of Logarithmic, Inverse Trigonometric Functions, and Implicit Differentiation

Derivatives of Logarithmic Functions

Basic Logarithmic Derivatives

The derivative of a logarithmic function depends on its base and argument. The most common logarithms in calculus are natural logarithms (base e) and common logarithms (base 10).

• Definition: For $y = \log_b x$, where $b$ is the base and $x$ is the argument:

• Derivative formulas:

  • $\frac{d}{dx}(\log_b x) = \frac{1}{x \ln b}$

  • $\frac{d}{dx}(\ln x) = \frac{1}{x}$

• Chain Rule for Logarithms: If $y = \ln g(x)$, then $y' = \frac{g'(x)}{g(x)}$

• General Rule: $\frac{d}{dx}(\ln g(x)) = \frac{g'(x)}{g(x)}$

Example: Differentiate $y = \ln(2x^4 + x)$

• $y' = \frac{(2x^4 + x)'}{2x^4 + x} = \frac{8x^3 + 1}{2x^4 + x}$

Example: Differentiate $f(x) = \log_5(4 + \cos x^3)$

• $y' = \frac{(4 + \cos x^3)'}{(4 + \cos x^3) \ln 5}$

• $(4 + \cos x^3)' = -\sin(x^3) \cdot 3x^2$

• $y' = \frac{-3x^2 \sin(x^3)}{(4 + \cos x^3) \ln 5}$

Example: Differentiate $y = \ln\left(\frac{x+1}{\sqrt{x-2}}\right)$

• Let $u = x+1$, $v = \sqrt{x-2}$

• $y' = \frac{\left(\frac{x+1}{\sqrt{x-2}}\right)'}{\frac{x+1}{\sqrt{x-2}}}$

• Apply quotient rule and simplify:

• $y' = \frac{(x+1)' \sqrt{x-2} - (x+1) \frac{1}{2}(x-2)^{-1/2} (x-2)'}{(x-2) (x+1)}$

• Final simplified form: $y' = \frac{(x-2) - \frac{1}{2}(x+1)}{(x-2) \sqrt{x-2}}$

• Additional info: The quotient rule and chain rule are used in this differentiation.

Derivatives of Inverse Trigonometric Functions

Basic Formulas

Inverse trigonometric functions have specific derivative formulas, often involving square roots and rational expressions.

Function	Derivative
$\frac{d}{dx}(\sin^{-1} x)$	$\frac{1}{\sqrt{1-x^2}}$
$\frac{d}{dx}(\cos^{-1} x)$	$-\frac{1}{\sqrt{1-x^2}}$
$\frac{d}{dx}(\tan^{-1} x)$	$\frac{1}{1+x^2}$
$\frac{d}{dx}(\sec^{-1} x)$	$\frac{1}{|x|\sqrt{x^2-1}}$
$\frac{d}{dx}(\csc^{-1} x)$	$-\frac{1}{|x|\sqrt{x^2-1}}$
$\frac{d}{dx}(\cot^{-1} x)$	$-\frac{1}{1+x^2}$

Chain Rule for Inverse Trigonometric Functions: If $y = f^{-1}(g(x))$, then $y' = f^{-1}'(g(x)) \cdot g'(x)$

• Example: Differentiate $y = 2 \sin^{-1} x$

  • Method 1: $y' = 2 \cdot \frac{1}{\sqrt{1-x^2}}$

  • Method 2 (chain rule): $y' = 2 [\sin^{-1} x]' = 2 \cdot \frac{1}{\sqrt{1-x^2}}$

• Example: Differentiate $y = \csc(x^2)$

  • $y' = -\csc(x^2) \cot(x^2) \cdot 2x$

• Example: Differentiate $y = \cos^{-1}(x^3 + 2)$

  • $y' = -\frac{1}{\sqrt{1-(x^3+2)^2}} \cdot 3x^2$

Function	Derivative (with chain rule)
$\cos^{-1}(f(x))$	$-\frac{1}{\sqrt{1-(f(x))^2}} \cdot f'(x)$
$\tan^{-1}(f(x))$	$\frac{1}{1+(f(x))^2} \cdot f'(x)$
$\csc^{-1}(f(x))$	$-\frac{1}{|f(x)|\sqrt{(f(x))^2-1}} \cdot f'(x)$
$\sec^{-1}(f(x))$	$\frac{1}{|f(x)|\sqrt{(f(x))^2-1}} \cdot f'(x)$
$\cot^{-1}(f(x))$	$-\frac{1}{1+(f(x))^2} \cdot f'(x)$

Additional info: The chain rule is essential when differentiating inverse trigonometric functions of composite arguments.

Implicit Differentiation

Concept and Steps

Implicit differentiation is used when a function is not given explicitly as $y = f(x)$, but rather as a relation involving both $x$ and $y$. This technique allows us to find $\frac{dy}{dx}$ even when $y$ cannot be isolated easily.

• Steps for Implicit Differentiation:

  1. Differentiating both sides of the equation with respect to $x$, treating $y$ as a function of $x$ (using the chain rule for terms involving $y$).

  2. Solving the resulting equation for $\frac{dy}{dx}$.

• Example: Differentiate $x^2 + y^2 = 4$

  • Differentiating both sides: $2x + 2y y' = 0$

  • Solving for $y'$: $y' = -\frac{x}{y}$

• Example: Use implicit differentiation to find $\frac{dy}{dx}$ for $x^2 - 4xy + y^2 = 4$

  • Step 1: Differentiate both sides: $2x - 4(xy)' + 2y y' = 0$

  • Apply product rule: $(xy)' = x y' + y$

  • Substitute: $2x - 4(x y' + y) + 2y y' = 0$

  • Expand: $2x - 4x y' - 4y + 2y y' = 0$

  • Group $y'$ terms: $-4x y' + 2y y' = 4y - 2x$

  • $y'(-4x + 2y) = 4y - 2x$

  • $y' = \frac{4y - 2x}{-4x + 2y}$

• Example: Find $y'$ if $x^3 + y^3 = 6xy$

  • Step 1: Differentiate both sides: $3x^2 + 3y^2 y' = 6(xy)'$

  • Product rule: $(xy)' = x y' + y$

  • Expand: $3x^2 + 3y^2 y' = 6x y' + 6y$

  • Group $y'$ terms: $3y^2 y' - 6x y' = 6y - 3x^2$

  • $y'(3y^2 - 6x) = 6y - 3x^2$

  • $y' = \frac{6y - 3x^2}{3y^2 - 6x}$

Graphical Interpretation: For equations like $x^2 + y^2 = 4$, the graph is a circle, not a function $y = f(x)$. Implicit differentiation allows us to find slopes of tangents at any point on the curve.

Additional info: Implicit differentiation is crucial for finding derivatives of curves not given in explicit form, such as circles, ellipses, and other relations.