Limits & Derivatives

LessonAbout 1.5 h1 of 2 in Differential Calculus

Differential Calculus: Limits and Derivatives

Limits of Functions and L’Hospital’s Rule

The limit of a function \(f(x)\) as \(x\) approaches a value \(a\) is the value that \(f(x)\) approaches as \(x\) gets arbitrarily close to \(a\). When direct substitution of \(x = a\) yields an indeterminate form such as \(0/0\) or \(\infty/\infty\), L’Hospital’s Rule allows evaluation by taking the derivative of the numerator and denominator separately.

L’Hospital’s Rule for Indeterminate Forms (\(0/0\) or \(\infty/\infty\)):
\[ \lim_{x \to a} \dfrac{f(x)}{g(x)} = \lim_{x \to a} \dfrac{f'(x)}{g'(x)} \]
Fundamental Trigonometric Limit:
\[ \lim_{x \to 0} \dfrac{1 - \cos x}{x^2} = \dfrac{1}{2} \]
Indeterminate Power Form (\(1^\infty\)):
\[ \text{If } L = \lim_{x \to a} [f(x)]^{g(x)} = 1^\infty, \qquad\qquad \text{then } \ln L = \lim_{x \to a} \dfrac{\ln f(x)}{1/g(x)} \]
  • Limits describe function behavior near specific input values
  • L’Hospital’s rule differentiates numerator and denominator separately
  • Indeterminate forms include zero over zero and infinity

Worked Example 1: Evaluation of an algebraic indeterminate limit

Evaluate \(\lim_{x \to 1} \dfrac{x^2 - 1}{x^2 + 3x - 4}\).

Solution: Direct substitution of \(x = 1\) yields \(\dfrac{0}{0}\). Applying L’Hospital’s rule by differentiating numerator and denominator separately:

\[ \lim_{x \to 1} \dfrac{\dfrac{d}{dx}(x^2 - 1)}{\dfrac{d}{dx}(x^2 + 3x - 4)} = \lim_{x \to 1} \dfrac{2x}{2x + 3} \]

Substitute \(x = 1\) into the simplified limit:

\[ \lim_{x \to 1} \dfrac{2(1)}{2(1) + 3} = \dfrac{2}{5} \]

Worked Example 2: Repeated application of L’Hospital’s rule

Evaluate \(\lim_{x \to 0} \dfrac{1 - \cos x}{x^2}\).

Solution: Direct substitution yields \(\dfrac{0}{0}\). Differentiating numerator and denominator:

\[ \lim_{x \to 0} \dfrac{\sin x}{2x} \]

Substituting \(x = 0\) still yields \(\dfrac{0}{0}\). Applying L’Hospital’s rule a second time:

\[ \lim_{x \to 0} \dfrac{\cos x}{2} = \dfrac{\cos 0}{2} = \dfrac{1}{2} \]

Worked Example 3: Limit at infinity

Compute the limit \(\lim_{x \to \infty} \dfrac{x + 4}{x - 4}\).

Solution: Direct evaluation gives \(\dfrac{\infty}{\infty}\). Apply L’Hospital’s rule:

\[ \lim_{x \to \infty} \dfrac{\dfrac{d}{dx}(x + 4)}{\dfrac{d}{dx}(x - 4)} = \lim_{x \to \infty} \dfrac{1}{1} = 1 \]

Worked Example 4: Indeterminate power form evaluation

Evaluate the limit \(\lim_{x \to 1} (2 - x)^{\tan(\frac{\pi x}{2})}\).

Solution: Direct substitution yields \(1^\infty\). Let \(L = \lim_{x \to 1} (2 - x)^{\tan(\frac{\pi x}{2})}\) and take the natural logarithm of both sides:

\[ \ln L = \lim_{x \to 1} \tan\left(\dfrac{\pi x}{2}\right) \ln(2 - x) = \lim_{x \to 1} \dfrac{\ln(2 - x)}{\cot\left(\dfrac{\pi x}{2}\right)} \]

This expression is in \(0/0\) form. Applying L’Hospital’s rule:

\[ \ln L = \lim_{x \to 1} \dfrac{\dfrac{-1}{2 - x}}{-\csc^2\left(\dfrac{\pi x}{2}\right) \cdot \dfrac{\pi}{2}} = \dfrac{\dfrac{-1}{2 - 1}}{-\csc^2\left(\dfrac{\pi}{2}\right) \cdot \dfrac{\pi}{2}} = \dfrac{-1}{-1 \cdot \dfrac{\pi}{2}} = \dfrac{2}{\pi} \]

Taking exponentials gives \(L = e^{2/\pi}\).

Derivatives of Algebraic, Exponential, and Logarithmic Functions

The derivative represents the instantaneous rate of change of a function with respect to an independent variable. Fundamental differentiation rules govern algebraic combinations, powers, radicals, exponentials, and logarithms.

Constant and Power Rules:
\[ \dfrac{d}{dx}(c) = 0, \qquad\qquad \dfrac{d}{dx}(u^n) = n u^{n-1} \dfrac{du}{dx} \]
Product and Quotient Rules:
\[ \dfrac{d}{dx}(uv) = u \dfrac{dv}{dx} + v \dfrac{du}{dx} \]
\[ \dfrac{d}{dx}\left(\dfrac{u}{v}\right) = \dfrac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]
Radical Rule:
\[ \dfrac{d}{dx}(\sqrt{u}) = \dfrac{1}{2\sqrt{u}} \dfrac{du}{dx} \]
Exponential Functions:
\[ \dfrac{d}{dx}(a^u) = a^u \ln a \dfrac{du}{dx}, \qquad\qquad \dfrac{d}{dx}(e^u) = e^u \dfrac{du}{dx} \]
Logarithmic Functions:
\[ \dfrac{d}{dx}(\ln u) = \dfrac{1}{u} \dfrac{du}{dx}, \qquad\qquad \dfrac{d}{dx}(\log_a u) = \dfrac{\log_a e}{u} \dfrac{du}{dx} \]
Variable Base and Exponent:
\[ \dfrac{d}{dx}(u^v) = v u^{v-1} \dfrac{du}{dx} + u^v \ln u \dfrac{dv}{dx} \]
  • Power rule multiplies exponent by base raised to reduced power
  • Product rule combines derivative of each factor individually
  • Quotient rule uses low d-high minus high d-low over low squared
  • Logarithmic differentiation simplifies complex exponent expressions

Worked Example 5: Product rule with exponential and trigonometric terms

Differentiate \(y = e^x \cos(x^2)\).

Solution: Apply the product rule \(\dfrac{d}{dx}(uv) = u v' + v u'\) where \(u = e^x\) and \(v = \cos(x^2)\):

\[ y' = e^x \dfrac{d}{dx}[\cos(x^2)] + \cos(x^2) \dfrac{d}{dx}(e^x) \]
\[ y' = e^x [-\sin(x^2) \cdot 2x] + \cos(x^2) e^x = e^x [\cos(x^2) - 2x \sin(x^2)] \]

Worked Example 6: Derivative of a logarithmic expression with base 10

Differentiate \(y = \log_{10}(x^2 + 1)^2\).

Solution: Simplify using logarithmic properties: \(y = 2 \log_{10}(x^2 + 1)\). Apply the base-10 logarithmic derivative formula \(\dfrac{d}{dx}(\log_{10} u) = \dfrac{\log_{10} e}{u} \cdot \dfrac{du}{dx}\):

\[ y' = 2 \left[ \dfrac{\log_{10} e}{x^2 + 1} \cdot (2x) \right] = \dfrac{4x \log_{10} e}{x^2 + 1} \]

Worked Example 7: Radical chain rule differentiation

Find the derivative with respect to \(x\) of \(y = \sqrt{2 - 3x^2}\).

Solution: Rewrite in power form \(y = (2 - 3x^2)^{1/2}\) and apply the radical rule:

\[ y' = \dfrac{1}{2\sqrt{2 - 3x^2}} \cdot \dfrac{d}{dx}(2 - 3x^2) = \dfrac{-6x}{2\sqrt{2 - 3x^2}} = \dfrac{-3x}{\sqrt{2 - 3x^2}} \]

Worked Example 8: Logarithmic differentiation for variable exponents

Find \(\dfrac{dy}{dx}\) for the implicit expression \((xy)^x = e\).

Solution: Take the natural logarithm of both sides:

\[ \ln[(xy)^x] = \ln e \implies x \ln(xy) = 1 \implies x(\ln x + \ln y) = 1 \]

Differentiate implicitly with respect to \(x\) using the product rule:

\[ 1 \cdot (\ln x + \ln y) + x \left( \dfrac{1}{x} + \dfrac{1}{y} \dfrac{dy}{dx} \right) = 0 \]
\[ \ln(xy) + 1 + \dfrac{x}{y} \dfrac{dy}{dx} = 0 \implies \dfrac{x}{y} \dfrac{dy}{dx} = -(1 + \ln(xy)) \]
\[ \dfrac{dy}{dx} = -\dfrac{y(1 + \ln(xy))}{x} \]

Derivatives of Trigonometric, Inverse Trigonometric, and Hyperbolic Functions

Trigonometric functions differentiate into co-functions or squared secants. Inverse trigonometric functions differentiate into algebraic fractions with square roots. Hyperbolic functions follow derivative forms similar to standard trigonometric functions.

Trigonometric Function Derivatives:
\[ \dfrac{d}{dx}(\sin u) = \cos u \dfrac{du}{dx}, \qquad\qquad \dfrac{d}{dx}(\cos u) = -\sin u \dfrac{du}{dx} \]
\[ \dfrac{d}{dx}(\tan u) = \sec^2 u \dfrac{du}{dx}, \qquad\qquad \dfrac{d}{dx}(\cot u) = -\csc^2 u \dfrac{du}{dx} \]
\[ \dfrac{d}{dx}(\sec u) = \sec u \tan u \dfrac{du}{dx}, \qquad\qquad \dfrac{d}{dx}(\csc u) = -\csc u \cot u \dfrac{du}{dx} \]
Inverse Trigonometric Derivatives:
\[ \dfrac{d}{dx}(\arcsin u) = \dfrac{1}{\sqrt{1-u^2}} \dfrac{du}{dx}, \qquad\qquad \dfrac{d}{dx}(\arccos u) = \dfrac{-1}{\sqrt{1-u^2}} \dfrac{du}{dx} \]
\[ \dfrac{d}{dx}(\arctan u) = \dfrac{1}{1+u^2} \dfrac{du}{dx}, \qquad\qquad \dfrac{d}{dx}(\text{arccot } u) = \dfrac{-1}{1+u^2} \dfrac{du}{dx} \]
Hyperbolic Function Derivatives:
\[ \dfrac{d}{dx}(\sinh u) = \cosh u \dfrac{du}{dx}, \qquad\qquad \dfrac{d}{dx}(\cosh u) = \sinh u \dfrac{du}{dx} \]
\[ \dfrac{d}{dx}(\tanh u) = \text{sech}^2 u \dfrac{du}{dx}, \qquad\qquad \dfrac{d}{dx}(\text{sech } u) = -\text{sech } u \tanh u \dfrac{du}{dx} \]
  • Derivative of sine equals cosine times derivative of angle
  • Derivative of cosine equals negative sine times angle derivative
  • Inverse trigonometric derivatives introduce algebraic square roots
  • Hyperbolic derivatives mirror standard trigonometric derivative forms

Worked Example 9: Secant composite differentiation

Differentiate \(y = \sec(x^2 + 2)\).

Solution: Apply the secant derivative rule with \(u = x^2 + 2\) and \(du/dx = 2x\):

\[ y' = \sec(x^2 + 2) \tan(x^2 + 2) \cdot \dfrac{d}{dx}(x^2 + 2) = 2x \sec(x^2 + 2) \tan(x^2 + 2) \]

Worked Example 10: Inverse cosine function derivative

Find the derivative of \(y = \arccos(4x)\).

Solution: Apply the arccosine rule with \(u = 4x\) and \(du/dx = 4\):

\[ y' = \dfrac{-1}{\sqrt{1 - (4x)^2}} \cdot \dfrac{d}{dx}(4x) = \dfrac{-4}{\sqrt{1 - 16x^2}} \]

Worked Example 11: Inverse sine of cosine function

Find \(y'\) if \(y = \arcsin(\cos x)\).

Solution: Set \(u = \cos x\), so \(du/dx = -\sin x\). Apply the arcsine rule:

\[ y' = \dfrac{1}{\sqrt{1 - \cos^2 x}} \cdot (-\sin x) = \dfrac{-\sin x}{\sqrt{\sin^2 x}} = \dfrac{-\sin x}{\sin x} = -1 \]

Worked Example 12: Trigonometric power reduction derivative

Find the derivative of \(y = 2\cos^2(x^2 + 2)\).

Solution: Apply the double-angle identity \(2\cos^2 \theta = 1 + \cos 2\theta\):

\[ y = 1 + \cos[2(x^2 + 2)] = 1 + \cos(2x^2 + 4) \]

Differentiate with respect to \(x\) using the chain rule:

\[ y' = -\sin(2x^2 + 4) \cdot (4x) = -4x \sin 2(x^2 + 2) \]

Implicit Differentiation and Higher-Order Derivatives

When an equation defines \(y\) implicitly as a function of \(x\), differentiation is performed term-by-term with respect to \(x\), applying the chain rule to terms containing \(y\). Partial derivatives measure rates of change with respect to one variable while holding all other variables constant.

Implicit Differentiation Chain Rule:
\[ \dfrac{d}{dx}[f(y)] = f'(y) \cdot \dfrac{dy}{dx} \]
Second Derivative Notation:
\[ y'' = f''(x) = \dfrac{d^2 y}{dx^2} = \dfrac{d}{dx}\left(\dfrac{dy}{dx}\right) \]
Partial Derivative with respect to \(x\):
\[ \dfrac{\partial f(x, y)}{\partial x} = \lim_{\Delta x \to 0} \dfrac{f(x + \Delta x, y) - f(x, y)}{\Delta x} \qquad\qquad (y = \text{constant}) \]
  • Implicit differentiation treats y as dependent function of x
  • Chain rule attaches y prime to every differentiated y term
  • Partial differentiation treats non-target variables as constants
  • Second derivative measures concavity and rate of slope change

Worked Example 13: Second derivative of an implicit curve

Find the second derivative \(y''\) by implicit differentiation from the equation \(4x^2 + 8y^2 = 36\).

Solution: Simplify the equation by dividing by 4: \(x^2 + 2y^2 = 9\). Differentiate both sides with respect to \(x\):

\[ 2x + 4y y' = 0 \implies y' = \dfrac{-2x}{4y} = \dfrac{-x}{2y} \]

Differentiate \(y'\) using the quotient rule to solve for \(y''\):

\[ y'' = -\dfrac{1}{2} \left[ \dfrac{y(1) - x y'}{y^2} \right] = -\dfrac{1}{2} \left[ \dfrac{y - x\left(\dfrac{-x}{2y}\right)}{y^2} \right] = -\dfrac{1}{2} \left[ \dfrac{2y^2 + x^2}{2y^3} \right] \]

Substitute \(x^2 + 2y^2 = 9\) into the numerator:

\[ y'' = -\dfrac{1}{2} \left[ \dfrac{9}{2y^3} \right] = -\dfrac{9}{4y^3} \]

Worked Example 14: Partial derivative evaluation

Find the partial derivative with respect to \(x\) of the function \(f(x, y) = x y^2 - 5y + 6\).

Solution: Treat \(y\) as a constant and differentiate with respect to \(x\):

\[ \dfrac{\partial}{\partial x}(x y^2 - 5y + 6) = y^2 \cdot \dfrac{d}{dx}(x) - 0 + 0 = y^2(1) = y^2 \]

Geometric Applications: Tangents, Normals, and Curvature

The first derivative gives the slope \(m\) of the tangent line to a curve at any point. The normal line is perpendicular to the tangent line. The second derivative determines concavity and curvature.

Slope of Tangent Line (\(m_1\)) and Normal Line (\(m_2\)):
\[ m_1 = y' = \left.\dfrac{dy}{dx}\right|_{(x_1, y_1)}, \qquad\qquad m_2 = -\dfrac{1}{m_1} \]
Equations of Tangent and Normal Lines:
\[ \text{Tangent: } y - y_1 = m_1(x - x_1), \qquad\qquad \text{Normal: } y - y_1 = m_2(x - x_1) \]
Radius of Curvature (\(R\)):
\[ R = \dfrac{[1 + (y')^2]^{3/2}}{|y''|} \]
Curvature (\(k\)):
\[ k = \dfrac{1}{R} = \dfrac{|y''|}{[1 + (y')^2]^{3/2}} \]
Critical Points (\(y' = 0\)) and Points of Inflection (\(y'' = 0\)):
\[ \text{Maximum: } y' = 0 \text{ and } y'' < 0 \qquad\qquad (\text{concave down}) \]
\[ \text{Minimum: } y' = 0 \text{ and } y'' > 0 \qquad\qquad (\text{concave up}) \]
\[ \text{Point of Inflection: } y'' = 0 \]
  • Slope of tangent line equals first derivative at point
  • Normal line is perpendicular to tangent line at point
  • Stationary points occur where first derivative equals zero
  • Points of inflection occur where second derivative equals zero

Worked Example 15: Tangent slope to a cubic curve

Find the slope of the tangent to the curve \(y = 2x - x^2 + x^3\) at the point \((0, 0)\).

Solution: Differentiate \(y\) to get the slope function \(y'\):

\[ y' = 2 - 2x + 3x^2 \]

Substitute \(x = 0\) into \(y'\):

\[ m = 2 - 2(0) + 3(0)^2 = 2 \]

Worked Example 16: Equation of the normal line to a circle

Find the equation of the normal to the circle \(x^2 + y^2 = 5\) at the point \((2, 1)\).

Solution: Differentiate implicitly to find tangent slope \(m_1\):

\[ 2x + 2y y' = 0 \implies y' = -\dfrac{x}{y} \]

At \((2, 1)\), tangent slope \(m_1 = -\dfrac{2}{1} = -2\). The normal line slope \(m_2\) is the negative reciprocal:

\[ m_2 = -\dfrac{1}{-2} = \dfrac{1}{2} \]

Using point-slope form \(y - y_1 = m_2(x - x_1)\):

\[ y - 1 = \dfrac{1}{2}(x - 2) \implies 2y - 2 = x - 2 \implies x = 2y \]

Worked Example 17: Critical points of a polynomial curve

Find the critical points of the curve \(y = 2 + 12x - x^3\).

Solution: Critical points occur where the slope \(y' = 0\):

\[ y' = 12 - 3x^2 = 0 \implies 3x^2 = 12 \implies x^2 = 4 \implies x = \pm 2 \]

Evaluate \(y\) at \(x = 2\) and \(x = -2\):

At \(x = 2\): \(y = 2 + 12(2) - (2)^3 = 2 + 24 - 8 = 18 \implies (2, 18)\).

At \(x = -2\): \(y = 2 + 12(-2) - (-2)^3 = 2 - 24 + 8 = -14 \implies (-2, -14)\).

The critical points are \((2, 18)\) and \((-2, -14)\).

Worked Example 18: Points of inflection location

Locate the points of inflection of the curve \(y = x^2 e^x\).

Solution: Find the first derivative \(y'\) using the product rule:

\[ y' = x^2 e^x + 2x e^x \]

Find the second derivative \(y''\) and set it to zero for points of inflection:

\[ y'' = (x^2 e^x + 2x e^x) + (2x e^x + 2e^x) = e^x (x^2 + 4x + 2) = 0 \]

Since \(e^x \neq 0\), solve \(x^2 + 4x + 2 = 0\) using the quadratic formula:

\[ x = \dfrac{-4 \pm \sqrt{4^2 - 4(1)(2)}}{2(1)} = \dfrac{-4 \pm \sqrt{8}}{2} = -2 \pm \sqrt{2} \]

Worked Example 19: Radius of curvature of a parabola

Find the radius of curvature of the parabola \(y^2 - 4x = 0\) at the point \((4, 4)\).

Solution: Differentiate \(y^2 - 4x = 0\) implicitly:

\[ 2y y' - 4 = 0 \implies y' = \dfrac{4}{2y} = \dfrac{2}{y} \]

Differentiate again to find \(y''\):

\[ y'' = \dfrac{d}{dx}\left(\dfrac{2}{y}\right) = \dfrac{-2 y'}{y^2} = \dfrac{-2(2/y)}{y^2} = \dfrac{-4}{y^3} \]

Substitute \(y = 4\) into \(y'\) and \(y''\):

\[ y' = \dfrac{2}{4} = 0.5, \qquad\qquad y'' = \dfrac{-4}{4^3} = \dfrac{-4}{64} = -\dfrac{1}{16} = -0.0625 \]

Substitute into the radius of curvature formula:

\[ R = \dfrac{[1 + (0.5)^2]^{3/2}}{|-0.0625|} = \dfrac{(1.25)^{1.5}}{0.0625} = \dfrac{1.39754}{0.0625} = 22.36\text{ units} \]

Worked Example 20: Radius of curvature of a logarithmic curve

Find the radius of curvature at any point of the curve \(y + \ln \cos x = 0\).

Solution: Express \(y\) explicitly: \(y = -\ln \cos x = \ln \sec x\). Differentiate to find \(y'\) and \(y''\):

\[ y' = -\dfrac{-\sin x}{\cos x} = \tan x \]
\[ y'' = \sec^2 x \]

Substitute \(y'\) and \(y''\) into the radius of curvature formula:

\[ R = \dfrac{[1 + (\tan x)^2]^{3/2}}{|\sec^2 x|} = \dfrac{(\sec^2 x)^{3/2}}{\sec^2 x} = \dfrac{\sec^3 x}{\sec^2 x} = \sec x \]

Formula Summary: Differential Calculus (Limits and Derivatives)

Formula Name Formula
L’Hospital’s Rule \(\lim_{x \to a} \dfrac{f(x)}{g(x)} = \lim_{x \to a} \dfrac{f'(x)}{g'(x)}\)
Fundamental Trigonometric Limit \(\lim_{x \to 0} \dfrac{1 - \cos x}{x^2} = \dfrac{1}{2}\)
Power Rule \(\dfrac{d}{dx}(u^n) = n u^{n-1} \dfrac{du}{dx}\)
Product Rule \(\dfrac{d}{dx}(uv) = u \dfrac{dv}{dx} + v \dfrac{du}{dx}\)
Quotient Rule \(\dfrac{d}{dx}\left(\dfrac{u}{v}\right) = \dfrac{v \dfrac{du}{dx} - u \dfrac{dv}{dx}}{v^2}\)
Radical Rule \(\dfrac{d}{dx}(\sqrt{u}) = \dfrac{1}{2\sqrt{u}} \dfrac{du}{dx}\)
Exponential Function Derivative \(\dfrac{d}{dx}(e^u) = e^u \dfrac{du}{dx}\)
General Exponential Derivative \(\dfrac{d}{dx}(a^u) = a^u \ln a \dfrac{du}{dx}\)
Natural Logarithm Derivative \(\dfrac{d}{dx}(\ln u) = \dfrac{1}{u} \dfrac{du}{dx}\)
General Logarithm Derivative \(\dfrac{d}{dx}(\log_a u) = \dfrac{\log_a e}{u} \dfrac{du}{dx}\)
Variable Base and Exponent Derivative \(\dfrac{d}{dx}(u^v) = v u^{v-1} \dfrac{du}{dx} + u^v \ln u \dfrac{dv}{dx}\)
Sine Derivative \(\dfrac{d}{dx}(\sin u) = \cos u \dfrac{du}{dx}\)
Cosine Derivative \(\dfrac{d}{dx}(\cos u) = -\sin u \dfrac{du}{dx}\)
Tangent Derivative \(\dfrac{d}{dx}(\tan u) = \sec^2 u \dfrac{du}{dx}\)
Secant Derivative \(\dfrac{d}{dx}(\sec u) = \sec u \tan u \dfrac{du}{dx}\)
Inverse Sine Derivative \(\dfrac{d}{dx}(\arcsin u) = \dfrac{1}{\sqrt{1-u^2}} \dfrac{du}{dx}\)
Inverse Cosine Derivative \(\dfrac{d}{dx}(\arccos u) = \dfrac{-1}{\sqrt{1-u^2}} \dfrac{du}{dx}\)
Inverse Tangent Derivative \(\dfrac{d}{dx}(\arctan u) = \dfrac{1}{1+u^2} \dfrac{du}{dx}\)
Hyperbolic Sine Derivative \(\dfrac{d}{dx}(\sinh u) = \cosh u \dfrac{du}{dx}\)
Hyperbolic Cosine Derivative \(\dfrac{d}{dx}(\cosh u) = \sinh u \dfrac{du}{dx}\)
Implicit Derivative Chain Rule \(\dfrac{d}{dx}[f(y)] = f'(y) \cdot \dfrac{dy}{dx}\)
Tangent Line Slope \(m_1 = \left.\dfrac{dy}{dx}\right|_{(x_1, y_1)}\)
Normal Line Slope \(m_2 = -\dfrac{1}{m_1}\)
Radius of Curvature \(R = \dfrac{[1 + (y')^2]^{3/2}}{|y''|}\)
Curvature \(k = \dfrac{|y''|}{[1 + (y')^2]^{3/2}}\)