Problem of the Day

1. 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) 1 B) g(x) D) 0

2. 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) 1 B) g(x) D) 0

3. Tangent Line Approximations y = f(x) (a, f(a)) 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.

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

5. Example Find the linearization of the function f(x) = √x + 3 at 
a = 1 and use it to approximate the numbers √3.98 
and √4.05 f(x) = (x + 3)½ f '(x) = ½(x + 3)-½ 1) Find f '(x)

6. Example Find the linearization of the function f(x) = √x + 3 at 
a = 1 and use it to approximate the numbers √3.98 
and √4.05 f(x) = (x + 3)½ f '(x) = ½(x + 3)-½ 1) Find f '(x) f(1) = √1 + 3 = 2 f '(1) = ½(1 + 3)-½ = ¼ 2) Find f(a) & f '(a)

7. Example Find the linearization of the function f(x) = √x + 3 at 
a = 1 and use it to approximate the numbers √3.98 
and √4.05 f(x) = (x + 3)½ f '(x) = ½(x + 3)-½ 1) Find f '(x) f(1) = √1 + 3 = 2 f '(1) = ½(1 + 3)-½ = ¼ 2) Find f(a) & f '(a) 3) Put in linear approximation form f(x) = f(a) + f '(a)(x - a) f(x) = 2 + ¼(x - 1) = 7/4 + ¼x

8. Example Find 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 form f(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

9. Example Find 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 form f(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

10. (c + Δx, f(c + Δx)) } (c, f(c)) f(c + Δx) { { f(c) c c + Δx Δx

11. y = f(c) + f '(c)(x - c) (c + Δx, f(c + Δx)) } } Δy } (c, f(c)) f '(c)dx f(c + Δx) { { f(c) c c + Δx Δx

13. Δy = the amount that the curve rises or falls dy = amount that the tangent line rises or falls as Δx changes, a change is produced in Δy as dx changes, a change is produced in dy

14. Differentials y = f(c) + f '(c)(x - c) (c + Δx, f(c + Δx)) } } Δy } (c, f(c)) f '(c)dx f(c + Δx) { { f(c) c c + Δx lim f(c + Δx) - f(c) = dy Δx 0 Δx dx f ' (c) = dy dx f '(c) dx = dy Δx when Δx is small we get the best approximation

15. Differentials Δy = f(x + Δx) - f(x) dy = f '(c)dx dy is the differential of y or Δy ≈ dy dx is the differential of x Δy = the amount that the curve rises or falls dy = amount that the tangent line rises or falls In many applications, the differential of y can be used 
as an approximation of the change in y.

16. If y = x2 then dy/dx = 2x dy = 2xdx and Δy = (x + Δx)2 - x2 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

17. If y = x2 then dy/dx = 2x dy = 2xdx and Δy = (x + Δx)2 - x2 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

18. If y = x2 then dy/dx = 2x dy = 2xdx Δy = (x + Δx)2 - x2 Δy = (x + Δx)2 - x2 Δy = (1 + 2)2 - 12 Δy = 9 - 1 Δy = 8 dy = 2xdx dy = 2(1)(2) = 4

19. Calculating Differentials Differential Derivative Function y = x2 2x 2xdx y = 2 sin x 2 cos x 2 cos x dx y = 1/x -1/x2 -dx/x2

20. Error propagation (Using differentials) Physicists and engineers use dy to approximate Δy. 
An example is estimating errors in measuring devices. { propagated error measurement error { { f(x + Δx) - f(x) = Δy { measured value exact value You can use differentials instead of calculating

21. 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

22. 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 is dV = 4πr2dr V 4/3πr3 dV = 4πdr V 4/3πr = 3(0.05) = .7% 21