Differentiation

Definition of the Derivative

The derivative of the function ƒ, or instantaneous rate of change, is given by converting the slope of the secant line to the slope of the tangent line by making the change is x, Δ(x) or h, approach 0.

(ƒ^′)(x) = (lim_h→0)((ƒ(x+h)-ƒ(x))/h)

Alternate Definition:

(ƒ^′)(c) =(lim_x→c)((ƒ(x)-ƒ(c))/(x-c))

Differentiability and Continuity Properties

A. If ƒ(x) is differentiable at x=c, then ƒ(x) is continuous at x=c

B. If ƒ(x) is not continuous at x=c, then ƒ(x) is not differentiable at x=c

C. The graph of f is continuous, but not differentiable at x=c if:

I. The graph has a cusp or sharp point at x=c

II. The graph has a vertical tangent line at x=c

III. The graph has an endpoint at x=c

Basic Derivative Rules

Given c is a constant:

Constant Rule: d(c)/d(x) = 0

Constant Multiple Rule: d([c*ƒ(x)])/d(x)=c*(ƒ^′)(x)

Sum Rule: d([ƒ(x)+g(x)])/d(x)=(ƒ^′)(x)+(g^′)(x)

Difference Rule: d([ƒ(x)-g(x)])/d(x)=(ƒ^′)(x)-(g^′)(x)

Product Rule: d([ƒ(x)*g(x)])/d(x)=(ƒ^′)(x)*g(x)+ƒ(x)*(g^′)(x)

Quotient Rule: d([ƒ(x)/g(x)])/d(x)=(g(x)*(ƒ^′)(x)-ƒ(x)*(g^′)(x))/([g(x)]2)

Chain Rule: d([ƒ(g(x))])/d(x)=(ƒ^′)(g(x))*(g^′)(x)

Derivatives of Trig Functions

d([sin(x)])/d(x)=cos(x)

d([cos(x)])/d(x)=-sin(x)

d([tan(x)])/d(x)=(sec^2)(x)

d([sec(x)])/d(x)=sec(x)*tan(x)

d([csc(x)])/d(x)=-csc(x)*cot(x)

d([cot(x)])/d(x)=-(csc^2)(x)

Derivatives of Trig Functions

d([(sin^-1)(x)])/d(x)=1/√(,1-x2)

d([(cos^-1)(x)])/d(x)=(-1)/√(,1-x2)

d([(tan^-1)(x)])/d(x)=1/(1+x2)

d([(sec^-1)(x)])/d(x)=1/(|x|√(,x2-1))

d([(csc^-1)(x)])/d(x)=(-1)/(|x|√(,x2-1))

d([(cot^-1)(x)])/d(x)=(-1)/(1+x2)

Derivatives of Exponential and Logarithmic Functions

(lim_x→0)(1+x)1/x=e

d([(log_a)(x)])/d(x)=1/(x*ln(a)), (a>0,a≠1)

d([ln(x)])/d(x)=1/x

d([ln(|x|)])/d(x)=1/x

d([(log_a)(|x|)])/d(x)=1/(x*ln(a)),(a>0,a≠1)

d([ex])/d(x)=ex

d([ax])/d(x)=ax*ln(a)

Explicit and Implicit Differentiation

A. Explicit Functions: Function y is written only in terms of the variable x (y=ƒ(x)). Apply derivatives rules normally.

B. Implicit Differentiation: An expression representing the graph of a curve in terms of both variables xand y

I. Differentiate both sides of the equation with respect to x (terms with x differentiate normally, terms with y are multiplied by d(y)/d(x)per the chain rule)

II. Group all terms with d(y)/d(x) on one side of the equation and all other terms on the other side of the equation.

III. Factory d(y)/d(x) and express d(y)/d(x) in terms of x and y

Tangent Lines and Normal Lines

A. The equation of the tangent line at a point (a,ƒ(a)):

y-ƒ(a) = (ƒ^′)(a)*(x-a)

B. The equation of the normal line at a point (a,ƒ(a)):

y-ƒ(a) = -1/(ƒ^′)(a)*(x-a)

Mean Value Theorem for Derivatives

If the function ƒ is continuous on the close interval [a,b] and differentiable on the open interval (a,b), then there exists at least one number c between a and b such that

(ƒ^′)(c) =(ƒ(b)-ƒ(a))/(b-a)

The slope of the tangent line is equal to the slope of the secant line

Rolle's Theorem (Special Case of Mean Value Theorem)

If the function ƒ is continuous on the close interval [a,b] and differentiable on the open internval (a,b), and ƒ(a) =ƒ(b), then there exists at least one number c between a and b such that

(ƒ^′)(c) = (ƒ(b)-ƒ(a))/(b-a) = 0

Particle Motion

A velocity function is found by taking the derivative of position. An acceleration function is found by taking the derivative of a velocity function.

Position: x(t)

Velocity: (x^′)(t) = v(t)

Speed: |v(t)|

Acceleration: (x^′*′)(t) = (v^′)(t) = a(t)

Rules:

I. If velocity is positive, the particle is moving right or up. If velocity is negative, the particle is moving left or down.

II. If velocity and acceleration have the same sign, the particle speed is increasing. If velocity and acceleration have opposite signs, speed is decreasing.

III. If velocity is zero and the sign of velocity changes, the particle changes direction.

Related Rates

A. Identify the known variables, including their rates of change and the rate of change that is to be found. Construct an equation relating the quantities whose rates of change are known and the rate of change to be found.

B. Implicitly differentiate both sides of the equation with respect to time. (Remember: DO NOT substitute the value of a variable that changes throughout the situation before you differentiate. If the value is constant, you can substitute it into the equation to simplify the derivative calculation).

C. Substitute the known rates of change and the known values of the variables into the equation. Then solve for the required rate of change.

*Keep in mind, the variables present can be related in different ways which often involves the use of similar geometric shapes, Pythagorean Theorem, etc.

Extrema of a Function

A. Absolute Extrema: An absolute maximum is the highest y – value of a function on a given interval or across the entire domain. An absolute minimum is the lowest y – value of a function on a given interval or across the entire domain.

B. Relative Extrema

Relative Maximum: The y-value of a function where the graph of the function changes from increasing to decreasing. Another way to define a relative maximum is the y-value where derivative of a function changes from positive to negative.

Relative Minimum: The y-value of a function where the graph of the function changes from decreasing to increasing. Another way to define a relative maximum is the y-value where derivative of a function changes from negative to positive.

Critical Value

When ƒ(c) is defined, if (ƒ^′)(c)=0 or (ƒ^′) is undefined at x=c, the values of the x – coordinate at those points are called critical values.

*If ƒ(x) has a relative extrema at x=c, then c is a critical value of ƒ

Extreme Value Theorem

If the function ƒ continuous on the closed interval [a,b], then the absolute extrema of the function ƒ on the closed interval will occur at the endpoints or critical values of ƒ

*After identifying critical values, create a table with endpoints and critical values. Calculate the y – value at each of these x values to identify the extrema.

Increasing and Decreasing Functions

For a differentiable function ƒ

A. If (ƒ^′)(x)>0 in (a,b), then ƒ is increasing on (a,b) →Tangent line has a positive slope

B. If (ƒ^′)(x) <0 in (a,b), then ƒ is decreasing on (a,b) → Tangent line has a negative slope

C. If (ƒ^′)(x) =0 in (a,b), then ƒ is constant on (a,b) → Tangent line has a zero slope (horizontal)

First Derivative Test

After calculating any discontinuities of a function ƒ and calculating the critical values of a function ƒ, create a sign chart for (ƒ^′), reflecting the domain, discontinuities, and critical values of a function ƒ.

A. If (ƒ^′)(x) changes sign from negative to positive at x=c, then ƒ(c) is a relative minimum of ƒ

B. If (ƒ^′)(x) changes sign from positive to negative at x=c, then ƒ(c) is a relative maximum of ƒ

*If there is no sign change of (ƒ^′)(x), there exists a shelf point

Concavity

For a differentiable function ƒ(x),

A. If (ƒ^″)(x)>0, the graph of ƒ(x) is concave up →(ƒ^′)(x) is increasing

B. If (ƒ^″)(x) <0, the graph of ƒ(x) is concave down →This means (ƒ^′)(x) is decreasing

Second Derivative Test

For a function ƒ(x) that is continuous at x=c

A. If (ƒ^′)(c)=0 and (ƒ^″)(c)>0, then ƒ(c) is a relative minimum.

B. If (ƒ^′)(c)=0 and (ƒ^″)(c)<0, then ƒ(c) is a relative maximum

* If (ƒ^′)(c)=0 and (ƒ^″)(c)=0, you must use the first derivative test to determine extrema

Point of Inflection

Let ƒ be a function whose second derivative exists on any interval. If ƒ is continuous at x=c, (ƒ^″)(c)=0 or (ƒ^″)(c) is undefined, and (ƒ^″)(x) changes sign at x=c, then the point (c,ƒ(c)) is a point of inflection.

Optimization

Finding the largest or smallest value of a function subject to some kind of constraints.

A. Define the primary equation for the quantity to be maximized or minimized. Define a feasible domain for the variables present in the equation.

B. If necessary, define a secondary equation that relates the variables present in the primary equation. Solve this equation for one of the variables and substitute into the primary equation.

C. Once the primary equation is represented in a single variable, take the derivative of the primary equation.

D. Find the critical values using the derivative calculated.

E. The optimal solution will more than likely be found at a critical value from D. Keep in mind, if the critical values do not represent a minimum or a maximum, the optimal solution may be found at an endpoint of the feasible domain.

Derivative of an Inverse

If ƒ and its inverse g are differentiable, and the point (c,ƒ(c)) exists on the function ƒ meaning the point (ƒ(c),c) exists on the function g, then

d([g(x)])/d(x)=1/(ƒ^′)((ƒ^-1)(x))=1/(ƒ^′)(ƒ(c))

BC Only: Derivatives of Parametric Functions

If ƒ and g are continuous functions of t on an interval, then the equations x=ƒ(t) and y=g(t) are called parametric equations, providing the position in the coordinate plane, and t is called the parameter.

A. The slope of the curve at the point (x,y) is

d(y)/d(x)=d(y)/d(t)/d(x)/d(t), provided d(x)/d(t)≠0

B. The second derivative at the point (x,y) is

(d^2)(y)/d(x2)=d(d(y)/d(x))/d(t)/d(x)/d(t)

Antiderivatives

If (F^′)(x)=ƒ(x) for all x, F(x) is an antiderivative of ƒ

(∫_^)(ƒ(x))=F(x)+C

* The antiderivative is also called the Indefinite Integral