Problem of the Day Let f and g be differentiable functions with the following properties:

(i) g(x) > 0 for all x(ii) f(0) = 1

If h(x) = f(x)g(x) and h'(x) = f(x)g'(x), then f(x) =

A) f '(x) C) ex E) 1B) g(x) D) 0

Problem of the Day Let f and g be differentiable functions with the following properties:

(i) g(x) > 0 for all x(ii) f(0) = 1

If h(x) = f(x)g(x) and h'(x) = f(x)g'(x), then f(x) =

A) f '(x) C) ex E) 1B) g(x) D) 0

(a, f(a))y = f(x)

The equation of the tangent line to the curve y = f(x) at (a, f(a)) is y - f(a) = f '(a)(x - a) or y = f(a) + f '(a)(x - a)

It is called the linear approximation of f at a and can be used to approximate values on the curve close to a.

Tangent Line Approximations

ExampleFind the linearization of the function f(x) = √x + 3 at a = 1 and use it to approximate the numbers √3.98 and √4.05

ExampleFind the linearization of the function f(x) = √x + 3 at a = 1 and use it to approximate the numbers √3.98 and √4.05

1) Find f '(x) f(x) = (x + 3)½f '(x) = ½(x + 3)-½

ExampleFind the linearization of the function f(x) = √x + 3 at a = 1 and use it to approximate the numbers √3.98 and √4.05

2) Find f(a) & f '(a) f(1) = √1 + 3 = 2f '(1) = ½(1 + 3)-½ = ¼

1) Find f '(x) f(x) = (x + 3)½f '(x) = ½(x + 3)-½

ExampleFind the linearization of the function f(x) = √x + 3 at a = 1 and use it to approximate the numbers √3.98 and √4.05

2) Find f(a) & f '(a) f(1) = √1 + 3 = 2f '(1) = ½(1 + 3)-½ = ¼

1) Find f '(x) f(x) = (x + 3)½f '(x) = ½(x + 3)-½

3) Put in linear approximation formf(x) = f(a) + f '(a)(x - a)f(x) = 2 + ¼(x - 1) = 7/4 + ¼x

ExampleFind the linearization of the function f(x) = √x + 3 at a = 1 and use it to approximate the numbers √3.98 and √4.05

3) Put in linear approximation formf(x) = 7/4 + ¼x

4) Approximate √3.98 (remember √3.98 is √.98 + 3, so what is x?)

√x + 3 = √3.98 ≈7/4 + ¼(.98)≈1.995

Now find √4.05

ExampleFind the linearization of the function f(x) = √x + 3 at a = 1 and use it to approximate the numbers √3.98 and √4.05

3) Put in linear approximation formf(x) = 7/4 + ¼x

4) Approximate √4.05 (remember √4.05 is √1.05 + 3, so what is x?)

√x + 3 = √4.05 ≈7/4 + ¼(1.05) ≈2.0125

c

(c, f(c))

c + Δx

f(c){

{Δx

(c + Δx, f(c + Δx))

} f(c + Δx)

}c

(c, f(c))

c + Δx

f(c){

{Δx

f '(c)dx

(c + Δx, f(c + Δx))

} Δy} f(c + Δx)

y = f(c) + f '(c)(x - c)

Δy = the amount that the curve rises or fallsdy = amount that the tangent line rises or falls

as Δx changes, a change is produced in Δyas dx changes, a change is produced in dy

}Differentials

c

(c, f(c))

c + Δx

f(c)

{

{Δx

f '(c)dx

(c + Δx, f(c + Δx))

} Δy}f(c + Δx)

y = f(c) + f '(c)(x - c)

when Δx is smallwe get the best approximation

lim f(c + Δx) - f(c) = dyΔx 0 Δx dx f ' (c) = dy dx f '(c) dx = dy

DifferentialsΔy = f(x + Δx) - f(x)

dy is the differential of y

dx is the differential of x

In many applications, the differential of y can be used as an approximation of the change in y.

or Δy ≈ dy

Δy = the amount that the curve rises or fallsdy = amount that the tangent line rises or falls

dy = f '(c)dx

Find dy when x = 1 and dx = 0.01. Compare this value with Δy for x = 1 and Δx = 0.01.

dy = f '(x)dx = f '(1)(0.01) = 0.02

Δy = f(x + Δx) - f(x) = f(1.01) - f(1) = 0.0201

If y = x2 then dy/dx = 2x dy = 2xdx

and Δy = (x + Δx)2 - x2

If y = x2 then dy/dx = 2x dy = 2xdx

If you start at x = 1 and move along the tangent line so that the change in x is 2 (i.e. dx and Δx), find dy and Δy

and Δy = (x + Δx)2 - x2

dy = 2xdx dy = 2(1)(2) = 4

Δy = (x + Δx)2 - x2

Δy = (x + Δx)2 - x2 Δy = (1 + 2)2 - 12Δy = 9 - 1Δy = 8

If y = x2 then dy/dx = 2x dy = 2xdx

Calculating Differentials

y = x2 2x 2xdx

y = 2 sin x 2 cos x 2 cos x dx

y = 1/x -1/x2 -dx/x2

Function Derivative Differential

Error propagation (Using differentials)

Physicists and engineers use dy to approximate Δy. An example is estimating errors in measuring devices.

f(x + Δx) - f(x) = Δy{exact value

{measurement error

{measured value

propagated error{

You can use differentials instead of calculating

A sphere is measured and found to be 21 cm with a possible measurement error of 0.05 cm. What is the maximum error in using this value of the radius to compute the volume of the sphere?

V = 4/3πr3(if error in r is dr or Δr then error in V is dV or ΔV)

dV = 4πr2dr = 4π(21)2(0.05) ≈277

Thus the maximum error in the calculated volume is 277 cm3

A sphere is measured and found to be 21 cm with a possible measurement error of 0.05 cm. What is the maximum error in using this value of the radius to compute the volume of the sphere?

The % error in volume isdV = 4πr2dr V 4/3πr3

dV = 4πdr V 4/3πr

= 3(0.05) = .7% 21