Differentiation 2.1-2.4

Differentiation 2.1-2.4 Taylor Woods, David Won, and McKenzie Brist

2.1- The Derivative and Tangent Line Problem • 2.2- Basic Differentiation Rules and Rates of Change • 2.3- Product and Quotient Rules and Higher Order Derivatives • 2.4- The Chain Rule Table of Contents

Definition of Tangent Line with Slope m: • If f is defined on an open interval containing c, and if the limit • Lim ∆x→0 ∆y/∆x= lim∆x→0 [(f(c+∆x) – f(c))/∆x] = m • Exists, then the line passing through (c,f(c)) with slope m is the tangent line to the graph of f at the point (c,f(c)). • Definition of the Derivative of a Function: • The derivative of f at x is given by • f’(x) = lim∆x→0 [(f(x+∆x) – f(x))/∆x] • provided the limit exists. For all x for which this limit exists, f’ is a function of x. • The process of finding a derivative is called differentiation. There are different notations for derivatives. They include : f’(x), dy/dx, y’, d/dx [f(x)], Dx[y] • Ex: Find the derivative of f(x) = x2+4x • f’(x) = lim∆x→0[((x+∆x)2+ 4(x+∆x))- (x2+4x)]/∆x • = lim∆x→0 [x2+ 2x∆x + ∆x2+ 4x + 4∆x – x2 – 4x]/∆x • = lim∆x→0 (2x∆x + ∆x2 + 4∆x)/∆x • =lim∆x→0 ∆x(2x +∆x + 4)/∆x • =lim∆x→0 2x + ∆x + 4 • =2x + 4 Section 2.1

Alternate form of Derivative: • f’(c) = limx→c f(x) – f(c) / x-c • this equation only exists if one sided limts, such as : • limx→c- f(x) – f(c) / x-c and limx→c+ f(x)-f(c) / x-c • exists and are equal to one another. This means that f is differentiable on a the closed interval [a,b]. • Differentiability Implies Continuity: If f is differentiable at x = c, then f is continuous at x = c. http://www1.teachertube.com/viewVideo.php?video_id=141017&title=Derivatives_and_the_Tangent_Line&vpkey=d9ed952caf here is a video on tangent line and differentiation using limits. Section 2.1 continued…

Constant Rule– derivative of a constant function f(x) = c is zero. Written out this looks like d/dx(c)= 0. • Power Rule (Constant Multiple Rule) – derivative of a function f(x) =axn when n is a rational number is equal to anx(n-1); a = constant • Ex: f(x)=2x4 + 3x2 • f’(x)=2(4)x(4-1) + 3(2)x(2-1) • f’(x)=8x3 + 6x • The slope of a graph at a certain point equals the derivative at that point. • To find the equation of a tangent line, use the formula y-y1 = m(x-x1) where m=slope, x1 and y1 equals the value of certain points on the graph of f(x). • Sum and Difference Rule- the derivative of f+g or f-g (2 differentiable functions) is the sum or difference of the derivatives of the 2 functions f and g. Therefore: [f(x) ±g(x)] = [f’(x) ± g’(x)] • Derivatives of sine and cosine functions - d/dxsinx = cosx and d/dxcosx = -sinx Section 2.2

Rates of change– change in distance/change in time. Movement upward (or to the right) is positive and movement downward (or to the left) is negative. • Using derivative to find velocity – the velocity function is the derivative of the position function V(t) = S’(t) . The units for velocity is distance/time (ie. feet/second) • The speed of an object is the absolute value of its velocity (this value for the speed cannot be negative) • http://www1.teachertube.com/viewVideo.php?video_id=141017&title=Derivatives_and_the_Tangent_Line&vpkey=d9ed952caf further in this video are explanations of the power rule and sum and difference rule. Section 2.2 continued…

The Product Rule: • d/dx [f(x)g(x)] = f(x)g’(x) + g(x)f’(x) • this means that the derivative of fg is the first function times the derivative of the second, plus the second function times the derivative of the first. • The quotient Rule: • d/dx [f(x)/g(x)] = (g(x)f’(x) – f(x)g’(x))/[g(x)]2, g(x)≠ 0 • this means that the derivative of f/g is given by the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominator. • Derivatives of Trigonometric Functions: • d/dx [tan x] = sec2x d/dx [cot x] = -csc2x • d/dx [sec x] = sex x tan x d/dx [csc x[ = -csc x cot x Section 2.3

Higher-Order Derivatives: • Position function: s(t) • Velocity function v(t) = s’(t) • Acceleration function a(t) =v’(t) = s”(t) *second derivatives are denoted by “ sign. A third derivative would be written as y”’, and a forth would be written as y4. • Ex: s(t)= x2 + 4x – 12 Find velocity • s’(t)= 2x + 4; therefore the equation of the velocity is v(t)= 2x+4 • http://archives.math.utk.edu/visual.calculus/2/product_rule.5/index.html link to an explanation of the product rule and example problems • http://archives.math.utk.edu/visual.calculus/2/quotient_rule.4/index.html link to an explanation of the quotient rule with example problems. Section 2.3 continued…

Chain Rule = Derivative of outer function evaluated at the inner function times the derivative of the inner function. • Formula: F'(x) = f'(g(x))g'(x) • Examples of Decomposing Functions to fit chain rule: • Functionf(u) g (u) • a) y= 1/ x+1 y=1/u x+1 • b) y= sin (2X) sin u 2x • c) tan^2 x u^2 tan x • GENERAL POWER RULE: • When y =un you find the derivative by: n x u(n-1) ( u’ ) • Ex. F(x) = (3x+1)4 Let u= 3x+1 and n = 4 • F’ (x) = 4 ( 3x+1 )3 ( 3 x+ 1) ‘ • F’ (x) = 4 (3x +1)3 (3) • F’ (x) = 12 (3x+1)3 Section 2.4

SIMPLIFYING DERIVATIVES: • EX. F(x)= x^4 [ 5x-1]^3 • F’ (x)= 4x^3 [5x-1]^3 + x^4 (3) (5x-1)^2 (5) use product rule and chain rule • F’ (x)= 4x^3 [5x-1]^3 + 15 x^4 (5x-1)^2 • F’ (x)= x^3 ( 5x-1) ^2 [ 4 (5x-1) +15x] factor out x^3 and (5x-1)^2 • F’ (x) = x^3 (5x-1)^2 [35x-4] solution • TRIG FUNCTIONS AND CHAIN RULE: Section 2.4 continued…

TRIG FUNCTIONS AND CHAIN RULE: • http://archives.math.utk.edu/visual.calculus/2/chain_rule.4/index.html here is a link that gives more examples on the chain rule. 2.4 continued…

Here is a chart that summarizes all of the derivatives that were previously covered.

The End!

Differentiation 2.1-2.4