AP CALC BC NOTECARDS

Second derivative test for local extrema

1.) If the first derivative of f at c=0 and the second derivative is negative, then f has a local max at c 2.) If the first derivative of f at c=0 and the second derivative is positive, then f has a local min at c. Example: Find the local extreme values of x^3-12x-5 The first derivative is 3(x^2-4) and the second derivative is 6x. The critical points are +/- 2. The second derivative at -2 is negative (max) and at 2 its positive (min).

Polynomial and rational functions

1.If f(x)=anx^x+a(n-1)x^(x-1)+...+a0 is any polynomial function and c is any real number then, lim as x->c f(x)= f(c)= anc^n+a(n-1)c^(n-1)+...+a0 2.If f(x) and g(x) are polynomials and c is any real number, then lim as x->c f(x)/g(x)=f(c)/g(c), provided that g(c)≠0

Antiderivative

A function F(x) is an antiderivative of a function f(x) if F'(x) = f(x) for all x in the domain of f. The process of finding an antiderivative is antidifferentiation. Example: F'(x) = 3x^2. f(x) = x^3

One-sided and two-sided limits

A function f(x)has a limit as x approaches c if and only if the right-hand and left-hand limits at c exist and are equal. Example: Lim as x->2+= 5 and Lim as x->2-=5, then the Lim as x->2= 5

The IVT for continuous functions

A function y=f(x) that is continuous on a closed interval [a,b] takes on every value between f(a) and f(b). In other words, if y0 is between f(a) and f(b), then y0= f(c) for some c in [a,b]. Example: Prove that f(x) at some point on the interval [1,3] = 0. f(x)= x^3-x-1 changes from negative to positive from x=1 to x=2. Therefore at some point c between 1 and 2 there is a value of f(c)=0.

Critical point

A point in the interior of the domain of a function f at which f'=0 or f' does not exist is a critical point of f Example: y=x^(2/3) as a critical point at the sharp turn

Point of inflection

A point where the graph of a function has a tangent line and where the concavity changes. Example: Find all the points of inflection for y=e^-2x. The second derivative is (e^(-x^2))(4x^2-2). The first part is always positive while the second part changes sign at +/- sqrt 1/2. The inflection points are (-sqrt 1/2, 1/sqrt e) and (sqrt 1/2, 1/sqrt e)

Limit

Assume f is defined in a neighborhood of c and let c and l be real numbers. The functions f has limit l as x approaches c if, given any positive number E, there is a positive number & such that for all x, 0 < | x-c | < &. f(x)L |&| lim as x approaches c f(x) = L. Example: lim as x approaches 0 of sinx/x = 1.

Quotient rule

At a point where v≠0, the quotient y=u/v of two differentiable functions is differentiable, and d/dx(u/v)=[v(du/dx)-u(dv/dx)]/v^2

Differential estimate of change

Differential Estimate of Change Let f(x) be differentiable at x = a. The approximate change in the value of f when x changes from a to a + dx is df = f '(a)dx. The radius r of a circle increases from a = 10m to 10.1 m. Use dA to estimate the increase in the circle's area A. Compare this estimate with the true change in A, and find the approximate error. Solution Since A = πr^2, the estimated increase is dA = A '(a) dr = 2πa dr = 2π(10)(0.1) = 2π m^2. The true change is ∆A = π(10.1)^2 - π(10) = (102.01 - 100)π = 2.01π m^2 The approximation error is ∆A - dA = 2.01π - 2π = 0.01 π m^2

Differentials

Differentials Let y = f(x) be a differentiable function. The differential dx is an independent variable. The differential dy is dy = f'(x) dx Example: Find the differential dy and evaluate dy for the given values of x and dx. a) y = x^5 + 37x, x = 1, dx = 0.01 b) y = sin 3x, x = π, dx= -0.02 c) x + y = xy, x=2, dx = 0.05 SOLUTION a) dy = (5x^4 + 37) dx. When x = 1 and dx = 0.01, dy = (5 +37)(0.01 = 0.42 b) dy = (3cos 3x) dx. When x = π and dx = -0.02, dy = (3cos 3π)(-0.02) = 0.06. c) d(x + y) = d(xy) dx + dy = xdy + ydx dy(1 - x) = (y - 1)dx dy = (y -1)dx/1-x When x = 2 in the original equation, 2 + y = 2y, so y is also 2. Therefore dy = (2-1)(0.05)/(1-2) = -0.05.

Free-fall constants (earth)

English Units: g=32 ft/sec^2, s=1/2(32)t^2=16t^2|s in ft| Metric Units: g=9.8 m/sec^2, s=1/2(9.8)t^2= 4.9t^2 |s in meters| Example The Position Equation (also known as the freefall formula) S = -16t2 + Vot + So this equation only works with the acceleration of gravity on the surface of the earth of 32 feet per second each second. The equation for any gravity and units is S = at2 + Vot + So. a=-32 ft/sec2 represents the acceleration due to gravity. Answer Integrate both sides of s''(t) = -32 with respect to t to get This results in s'(t) = -32t + C1 where C1 is a constant and s'(t) is the velocity at any time t. At time t=0, s'(t) is equal to the initial velocity Vo so we get C1 = Vo. The new equation is s'(t) = -32t + Vo Now, integrate both sides with respect to t to get This results in where C2 is a constant and s(t) is the position at any time t. At time t=0, s(t) is equal to the initial position So so we get C2 = So. The final equation is or simplified

Power rule for positive integer powers of x

If "n" is a positive integer, then (d/dx)(x^n)=nx^(n-1) Exmaple: d/dx(x^2)= 2x^1, d/dx(x^3)=3x^2

Sum and difference rule

If "u" and "v" are differentiable functions of x, then their sum and difference are differentiable at every point where "u" and "v" are differentiable. At such points, (d/dx)(u±v)=(du/dx)±(dv/dx) Example:Find dp/dt if p=t^3+6t^2-(5/3)t+16 Sum and difference Rule: dp/dt=(d/dt)(t^3)+(d/dt)(6t^2)-(d/dt)(5/3)t+(d/dt)(16)

Constant multiple rule

If "u" is a differentiable function of x and c is a constant, then (d/dx)(cu)= c(du/dx) Example: d/dx(7x^4)= 7(4x^3)= 28x^3

Properties of limits as x->infinity

If L,M, and k are real numbers and limx->Infinity f(x)=L and limx->Infinity g(x)=M , then 1) Sum rule: limx->Infinity (f(x) + g(x))= L+M 2) Difference Rule: limx->Infinity (f(x)-g(x))=L-M 3) Product Rule: limx->Infinity (f(x)*g(x))=L*M 4) Constant Multiple Rule: limx->Infinity (k*f(x))=k*L 5) Quotient Rule: limx->Infinity f(x)/g(x)= L/M, M not 0 6) Power Rule: If r and s are integers, S is not 0, then limx->Infinity (f(x))^(r/s)= L^(r/s) provided that L^(r/s) is a real number Example: 1. Find Limx->Infinity (5x+ sinx)/(x) (5x + sinx)/(x) = 5x/x+ sinx/x = 5+ sinx/x limx->Infinity 5x+sinx/x = limx-> Infinity 5 + limx-> Infinty sinx/x = 5+0= 5

Local extreme values theorem

If a Function f has a local maximum value or a local minimum value at an interior point c of its domain, and if f' exists at c, then f'(c)=0 Example: -x^2 has a max at (0,0) and f'(0)=0

Finding dy/dx parametrically

If all three derivatives exist and dx/dt=/= 0, dy/dx= (dy/dt)/(dx/dt) Example: **Graph needed (pg. 151) Find tangent line to right-hand hyperbola branch defined parametrically by x= sec(t) y=tan(t) -π/2 < t < π/2 at the point (sqrt(2),1), where t=π/4 All 3 points exist, therefore dy/dx= (dy/dt)/(dx/dt) = sec^2(t)/sec(t)tan(t) =sec(t)/tan(t) =csc(t) Setting t=π/4 gives, =csc(π/4)= sqrt(2) Equation of tan. line is y-1=sqrt(2)(x-sqrt(2)) y=sqrt(2x)-2+1 =sqrt(2x)-1

Derivatives of inverse functions

If f is at every point on an interval I and (df)/(dx) is never zero on I, then f has an inverse amd f^-1 is differentiable at every point on the interval f(I). Example y = (1/2) x - 1 x = 2y + 2 y = 2x + 2 dy / dx = 2

Composite of continuous functions

If f is continuous at c and g is continuous at f(c), then the composite g of f is continuous at c. Example: both are continuous- f(x)= xsinx/ x^2+2 and g(x)= |x| so g(f(x))= |xsinx/ x^2+2 | is continuous too.

Extreme value theorem

If f is continuous on a closed interval [a,b], then f has both a maximum value and a minimum value on the interval Example: Maximum of h(x) is at (x2,M) and the minimum is at (x1,m)

Chain rule

If f is differentiable at the point u=g(x) and g is differentiable at x, then the composite function (f o g)(x)= f(g(x)) is differentiable at x, and (f o g)'(x)=f'(g(x)) * g'(x) Example: d/dx sin(x^2+x)= cos(x^2+x) * (2x+1)

Derivative of a constant function

If f is the function with the constant value "c", then: (df/dx)=(d/dx)(c)=0 Example: Find dp/dt if p=t^3+6t^2-(5/3)t+16 dp/dt=(d/dt)(t^3)+(d/dt)(6t^2)-(d/dt)(5/3)t+(d/dt)(16) Constant Rule: 3t^2+6(2t)-(5/3)+0

Functions with the same derivative differ by a constant

If f'(x) = g'(x) at each point of an interval I, then there is a constant C such that f(x) = g(x) + C for all x in I. Example: Find the function f(x) whose derivative is sin(x) and whose graph passes through the point (0,2). f(x) = -cos(x) + C. f(0) = 2. -cos(0) + C = 2. -1 + C = 2. C=3. f(x)= -cos(x) + 3

Functions with f´= 0 are constant

If f´(x)=0 at each point of an interval I, then there is a constant C for which f(x)=C for all x in I. Example: Since f is differentiable at every point of [X1,X2], f satisfies the hypotheses of the Mean Value Theorem. Therefore, there is a point c between X1 and X2 for which f´(c) = (f(X2) - f(X1)) / (X2 - X1). Because f´(c) = 0, it follows that f(X1) = f(X2).

Sandwich theorem

If g(x) <= f(x) <= h(x) for all x =/ c in some interval about c, and lim as x approaches c of g(x) = lim as x approaches h(x) = L then lim as x approaches 0 of f(x)= L. Example: Lim as x approaches 0 of x^2 sin (1/x) = |x^2 sin (1/x)|= -x^2 <= x^2 sin (1/x) <= x^2 = 0 <=x^2 sin (1/x)<=0 so the li as x approaches 0 of x^2 sin (1/x)= 0.

Power rule for negative integer powers of x

If n is a negative integer and x≠0, then d/dx(x^n)=nx^(n-1)

Power rule for rational powers of x

If n is any rational number, then d/dx x^n=nx^(n-1) If n < 1, then the derivative does not exist at x=0 Example: d/dx(sqrt(x))= d/dx(x^1/2)=1/2x^(-1/2)=1/2(sqrt(x)) *Notice that sqrt(x) is defined at x=0, but 1/2(sqrt(x)) is not.

Properties of continuous functions

If the functions f and g are continuous at x=c, then the following combinations are continuous at x=c. 1.Sums: f + g. 2.Differences: f - g. 3.Products: f • g. 4.Constant multiples: k • f, for any number k. 5.Quotients: f/g, provided g(c) ≠ 0. Example: Since f(x) = x^2 and g(x) = x^3 are both continuous at x = 2, then the sum of these functions, y = x^2 + x^3, is also continuous at x = 2.

Power rule for arbitrary real powers

If u is positive differentiable function of x and n is any real number, then u^n isa differentiable function of x, and (d/dx)u^n=nu^(n-1)(du/dx). Example If y=(2+sin3x)^π, then (d/dx)(2+sin3x)^π=π(2+sin3x)^(x-1)(cos3x)3=3π(2+sin3x)^(π-1)(cos3x)

Continuity at a point

Interior point: A function y=f(x) is continuous at an interior point c of its domain if the lim as x approaches c of f(x)= f(c). Endpoint: A function y=f(x) is continuous at a left endpoint a or is continuous at a right endpoint b of its domain if the lim as x approaches a+ of f(x)= f(A) or the lim as x approaches b- of f(x)= f(b). Example: lim as x approaches 3- of int x=2 and lim as x approaches 3+ int x= 3. The function is not continuous.

Local extreme values

Let c be an interior point of the domain of the function f. Then f(c) is a (a) local maximum value at c if and only if f(x)<_f(c) for all x in some open interval containing c. (b) Local Minimum value at c if and only if f(x)>_f(c) for all x in some open interval containing c. a function f has a local maximum or local minimum at an end point c if the appropriate inequality holds for all x in some Half-open domain interval containing c. Example: Local Maximum of h(x) is at (x2,M) and the Local minimum is at (x1,m)

Increasing function, decreasing function

Let f be a function defined on an interval I and let X1 and X2 be any two points in I. 1. f increases on I if X1 < X2 => f(X1) < f(X2). 2. f decreases on I if X1 < X2 => f(X1) > f(X2). Example: On [0,4], f(x)=x is increasing because point (1,1) is less than point (2,2).

Absolute extreme values

Let f be a function with domain D. Then f(c) is the (a) absolute maximum value on D if and only if f(x) ≤ f(c) for all x in D. (b) absolute minimum value on D if and only if f(x) ≥ f(c) for all x in D. Example: On [-π/2, π/2], f(x)=cos x takes on a maximum value of 1 once and a minimum value of 0 twice.

Increasing and decreasing functions

Let f be continuous on [a,b] and differentiable on (a,b). 1. If f´ > 0 at each point of (a,b), then f increases on [a,b]. 2. If f´< 0 at each point of (a,b), then f decreases on [a,b]. Example: The function y=x^2 is increasing on [0,∞) because y´= 2x > 0 on (0,∞).

Properties of limits

Lim as x >c f(x)= L and lim as x>c g(x)=M 1. Sum Rule: The limit of the sum of 2 functions is the sum of their limits. Lim as x>c (f(x)+g(x)= L+M 2. Difference Rule: The limit of the difference of two functions is the difference of their limits. Lim as x>c (f(x)-g(x))= L-M 3. Product Rule: The limit of a product of 2 functions is the product of their limits. Lim x>c (f(x) * g(x))= L *M 4. Constant Multiple Rule: The limit of a constant times a function is the constant ties the limit of the function. Lim x>c (k*f(x))= k*L 5. Quotient Rule: The limit of a quotient of two functions is the quotient of their limits, provided the limit of the denominator is not zero. Lim x>c f(x)/g(x)= L/M, M=/=0 6. Power Rule: If R and S are integers, s =/= 0, then Lim x>c (f(x))^(R/S)= L^(R/S) Example: Assume that Lim x>b f(x)=7 and Lim x>b g(x)=3 (a). Lim x>b (f(x)+g(x))= 7+3=10

Linearization

Linearization If f is differentiable at x = a, then the equation of the tangent line, L(x) = f (a) + f '(a)(x - a), defines the linearization of f at a. The approximation f (x) ≈ L(x) is the standard linear approximation of f at a. The point x = a is the center of the approximation. Example Find the linearization of f(x) = √(1+x) at x = 0, and use it to approximate √1.02 without a calculator. Then use a calculator to determine the accuracy of the approximation. SOLUTION Since f(0) = 1, the point of tangency is (0,1). Since f'(x) = ½(1 + x) ^ (-1/2 ) the slope of the tangent line is f'(0) = ½. Thus L(x) = 1 + ½ (x-0) = 1 x/2 To approximate √1.02, we use x = 0.02 √1.02 = f(0.02) ≈ L (0.02) = 1 + 0.02/2 = 1.01 The calculator gives √ 1.02 = 1.009950494, so the approximation error is |1.009950494 -1.01| ≈ 4.05 X 10^-5. The error is less than 10^-4.

Maximum profit

Maximum profit (if any) occurs at a production level at which marginal revenue equals marginal cost. Proof. We assume that r(x) and c(x) are differentiable for all x > 0, so if p(x) = r(x) - c(x) has a maximum value, it occurs at a production level at which p'(x) = 0. Since p'(x) = r'(x) - c'(x), p'(x) 0 implies that r'(x) - c'(x) = 0 or r'(x) = c'(x) Example: Suppose that r(x) = 9x and c(x) = x^3 - 6x^2 + 15x, where x represents thousands of units. Is there a production level that maximizes profit? If so, what is it? SOLUTION Notice that r'(x) = 9 and c'(x) = 3x^2 - 12x + 15. 3x^2 - 12x + 15 = 9 Set c'(x) = r'(x). 3x^2 -12x + 6 = 0 The two solutions of the quadratic equation are x1 = (12- √72)/6 = 2 -√2 ≈ 0.586 and x2 = (12+√72)/6 = 2 + √2 ≈ 3.414

Speed

The absolute value of velocity. Speed= |v(t)| = |ds/dt| Example A student walks around in front of a motion detector that records her velocity at 1 second intervals for 36 seconds. She stores the data in her graphing calculator and uses it to generate the time velocity graph. When is her speed a maximum? (pg 129 graph 3.25) Answer The student moves forward for the first 14 seconds, moves backwards for the next 12 seconds, stands still for 6 seconds, and then moves forward again. She achieves her max speed at t=20, while moving backward

Jerk

The derivative of acceleration. If a body's positions at time T is s(T)m the body's jerk time T is j(T)= da/dt=d^3s/dt^3 Example: Find the jerk time at time T if s=2cos(t) ds/dt= 2-sin(t) d^2s/dt^2= 2-cos(t) d^3s/dt^3= 2sin(t)=j(T)

Derivative

The derivative of the function f with respect to the variable x is the function f' whose value at x is f'(x)= lim as h->0 (f(x+h)-f(x))/ h provided the limit exists. Example: f(x)=x^3 (1)f'(x)=lim as h->0 ((x+h)^3-x^3)/h (2)= lim as h->0 ((x^3+3x^2h+3xh^2+h^3)-x^3)/h (3)=lim as h->0 ((3x^2+3xh+h^2)h)/h (4)f'(x)= 3x^2

Instantaneous velocity

The derivative of the position function s=f(t) with respect to time. At time t the velocity is v(t)=ds/ds= lim(delta t)->0 {f(t+ delta t)-f(t)}/delta t Example Let the following be the equation of motion: s(t) = 6t 2 + t + 8. Let t be measured in minutes and s in meters. And let 10:05 AM correspond to t = 0. What is the instantaneous velocity at 10:05 AM? Answer. ds/dt = 12t + 1 = 12· 0 + 1 = 1 m/min.

Acceleration

The derivative of velocity with respect to time. If a body's velocity at time t is v(t)=ds/dt then the body's acceleration at time t is a(t)=dv/dt= d^2s/dt^2 Example A particle moves along a line so that its position at any time t (greater than or equal to)0 is given by the function s(t)=t^2-4t+3, where s is measured in meters and t is measured in seconds Find the acceleration when the particle is at t=4 Answer The acceleration a(t) at any time t Is a(t)= dv/dt= 2m/sec^2. So a(4)=2

Inverse function-inverse identities

The derivatives of the inverse cofunctions are the negatives of the derivatives of the corresponding inverse functions. cos^(-1)x=(π/2)-sin^(-1)x cot^(-1)x=(π/2)-tan^(-1)x csc^(-1)x=(π/2)-sec^(-1)x

First derivative test for local extrema

The following test applies to a continuous function f(x). At a critical point c: 1) If f' changes sign from positive to negative at c (f'>0 for x<c and f'<0 for x>c), then f has a local maximum value at c. 2) If f' changes sign from negative to positive at c (f'<0 for x<c and f'>0 for x>c), then f has a local minimum value at c. 3) If f' does not change sign at c (f' has the same sign on both sides of c), then f has no local extreme value at c. At a left endpoint a: If f'<0 (f'>0) for x>a, then f has a local maximum (minimum) value at a. At a right endpoint b: If f'<0 (f'>0) for x<b, the f has a local minimum (maximum) value at b. Example: f(x)= x^3 - 12x - 5. f'(x)= 3x^2 - 12. f(x) = 0. x=2 and x=-2. Local max at x=-2 and local min at x=2. Local max value is f(-2) = 11 and the local min value is f(2) = -21

End behavior model

The function g is a. A right end behavior model for f if and only if lim as x approaches infinity f(x)/g(x)=1. b. A left end behavior model for f if and only if lim as x approaches negative infinity f(x)/g(x)=1. Example: f(x)=3x^4-2x^3+32^2-5x+6 g(x)= 3x^4 Lim as x approaches negative/positive infinity is (3x^4-2x^3+32^2-5x+6)/ 3x^4 = 1.

Concavity

The graph of a differentiable function y=f(x) is: a.) concave up on an open interval "I" if its first derivative is increasing on "I" b.) concave down on an open interval "I" if its first derivative is decreasing on "I"

Concavity test

The graph of a twice-differentiable function y=f(x) is concave up on any interval where the second derivative of y > 0 and concave down where the second derivative of y <0. Example: y=x^2 on (3,10) is always concave up because its second derivative is 2 and always positive.

Vertical asymptote

The line x=a is a vertical asymptote of the graph of a function y=f(x) if either lim as x approaches a+ f(x)= +- infinity or lim as x approaches a- f(x)=+- infinity. Example: lim as x approaches 0+ of 1/x^2= infinity and lim as x approaches 0- of 1/x^2= infinity so x=0 is a vertical asymptote.

Horizontal asymptote

The line y=b is a horizontal asymptote of the graph of a function y=f(x) is either lim as x approaches infinity of f(x)=b or the lim as x approaches negative infinity of f(x) = b. Example: f(x)= 2+1/x. Lim as x approaches infinity of 2 + 1/x = 2 the lim as x approaches negative infinity of 2+1/x= 2. Therefore the asymptote is at 2.

Product rule

The product of two differentiable functions u and v is differentiable, and d/dx (uv)= (u)dv/dx+(v)du/dx

Minimizing average cost

The production level (if any) at which average cost is smallest is a level at which the average cost equals the marginal cost. Proof. We assume that c(x) is differentiable. c(x) = cost of producing x items, x > 0 (c(x))/x = average cost of producing x items. If the average cost can be minimized, it will be a production level at which d/(dx ) ((c(x))/x)=0 (xc'(x)-c'(x))/x^2 = 0 xc'(x) - c(x) = 0 c'(x) = (c(x))/x Example: Suppose c(x) = x3 - 6x2 + 15x, where x represents thousands of units. Is there a production level that minimizes average cost? If so, what is it? SOLUTION. We look for levels at which average cost equals marginal cost. Marginal Cost: c'(x) = 3x2 - 12x + 15 Average Cost: x^2 - 6x + 15 3x2 - 12x +15 = x^2 - 6x +15 2x^2 - 6x = 0 2x(x - 3) = 0 x= 0 or x = 3

Instantaneous rate of change

The rate of change of f with respect to x at a is the derivative f'(a)=lim h->0 [f(a+h)-f(a)]/h , provided the limit exists.

Slope of a curve at a point

The slope of the curve y=f(x) at the point P(a, f(a)) is the number m= lim as h approaches 0 of f(a+h)-f(a)/ h. Example: f(x)= 4-x^2, at x=1. The lim as h approaches 0 of f(1+h)-f(1)/h= the lim as h approaches 0 of 4-(1+h)^2-3/h= the lim as h approaches 0 of 4-1-2h-h^2-3/h=the lim as h approaches 0 of -h(2+h)/h= -2.

Calculator conversion identities

We do not use tan^(-1)(1/x) as an identity for cot^(-1)x. A glance at the graphs of y=tan^(-1)(1/x) and y=(π/2)-tan^(-1)x reveals the problem. We cannot replace cot^(-1)x by the function y=tan^(-1)(1/x) in the identity for the inverse functions and inverse cofunctions, and so it is not the function we want for cot^(-1)x. sec^(-1)x=sin^(-1)(1/x) cot^(-1)x=(π/2)-tan^(-1)x csc^(-1)x=sin^(-1)(1/x)

IVT for derivatives

if a and b are any two points on an interval in which f is differentiable, then f' takes on every value between f'(a) and f'(b). Example: Show that f(x)=2x^3-5x^2-10x+5 has a root somewhere on the interval [-1,2]. Use f(x)=0 f(-1)=?= 8 f(2)=?= -19 so since f(2)<0<f(-1) we can say that the IVT is applicable here.

Differentiability implies continuity

if f has a derivative at x=a, then f is continuous at x=a

Mean value theorem for derivatives

if y=f(x) is continuous at every point of the closed interval [a.b] and differentiable at every point of its interior (a,b), then there is at least one point c in (a,b) at which f'(c)= (f(b)-f(a))/(b-a) Example:2c=f(2)-f(0)/2-0=2, c=1

Derivative at a point

the derivative of the function f at the point x=a is the limit f'(a)= lim as x->a f(x)-f(a)/x-a provided the limit exists. Example: f(x)=1/x a=2 f'(x)= lim as x->2(1/x-1/a)/x-a