Topic 8: Differentiation Flashcards
(31 cards)
What is the definition of differentiability (2,1)
-Suppose f: ℝ -> ℝ is defined at all points in a neighbourhood (open interval containing a) of x = and the limit lim x-> a (f(x) - f(a))/(x-a)
-Then we say f is differentiable at x = a and define f’(a) = limx->a to be the derivative of f at x = a
-If f(x) is defined for all x ∈ (a, b), then it is differentiable over (a,b) if it is differentiable at all x ∈ (a,b)
How can we prove that differentiable functions are continuous (4)
-f’(a) = limx->a (f(x) - f(a))/x-a
-limx->a(x-a)f’(a) = limx->af(x) - f(a)
-limx->a (f(x) - f(a)) = 0 -> limx->a f(x) = f(a)
-So F is continuous at x = a
What is the relationship between being differentiable and continuous in (a,b) (1)
-If f is differentiable in (a,b), its continuous in (a,b)
How to prove the derivative of sin(x) (7)
-Consider limx->a (sinx - sin(a))/(x-a)
-Use sin(c) - sin(D) = 2cos((c+d)/2)sin((c+d)/2)
-Hence (sin(x) - sin(a))/(x-a) = (2cos((x+a)/2)sin((x+a)/2))/(x-a)
=(cos((x+a)/2)sin((x-a)/2))/(0.5(x-a))
-As x -> a, (x-a) -> 0, so sin((x-a)/2))/(0.5(x-a)) -> 1
-Hence, limx->a (sinx - sin(a))/(x-a) = limx->a cos((x+a)/2) = cos a
-So the derivative of sin(x) is cos(x)
What is Rolle’s theorem (3,1)
Suppose f: ℝ -> ℝ is:
-Continuous in a closed interval [a,b]
-Differentiable in the open interval (a, b)
-f(a) = f(b)
-Then there exists c ∈ (a, b) such that f’(c) = 0
How can we apply Rolle’s theorem to find roots (3)
-Suppose we have function f: ℝ -> ℝ that is continuous and differentiable, either everywhere or in some interval (a,b)
-Suppose α, β ∈ ℝ are 2 solutions of f(x) = 0
-Then if the conditions of Rolle’s theorem are satisfied in (α, β), there must be a turning point c ∈ (α, β) such that f’(c) = 0
What do we know about a continuous and differentiable function with 2 roots (1)
-It must also contain a turning point
What do we know about functions where f’(x) ≠ 0 for all (a, b) (2)
-(a, b) contains at most one root
-If there were more than one, there would also be a turning point
What is the mean value theorem (2,1)
Suppose that f: ℝ -> ℝ is:
-Continuous in [a, b]
-Differentiable in (a, b)
-Then there exists c ∈ (a, b) such that (f(b) - f(a)/(b-a) = f’(c)
How can we prove the mean value theorem (8)
-Consider the function f: ℝ -> ℝ, given by F(X) = (b-a)(f(b) - f(x)) - (b-x)(f(b) - f(a))
-Note F(a) and F(b) = 0
-Furthermore, F is continuous in [a, b] and differentiable in (a, b) because f is
-Applying Rolle’s theorem, we know that there exists c ∈ (a, b) such that F’(c) = 0
-F’(x) = -(b-a)f’(x) +f(b) - f(A)
-Hence, F’(c) = 0 => -(b-a)f’(c) + f(b) - f(a) = 0
=> F’(c) = (f(b)-f(a))/(b-a) as claimed
What is a corollary of the mean value theorem (2)
-Say F is continuous on [a, b], and differentiable on (a, b) and f’(x) = 0
-For all x ∈ (a, b), then f(x) is constant for all x ∈ [a, b]
What is a corollary from 2 functions’ differentiated being equal (2)
-Say f, g: ℝ -> ℝ are continuous over [a, b] and differentiable over (a, b), such that f’(x) - g’(x) = 0 for all x ∈ (a, b)
-Then F(x) - g(x) is constant for all x ∈ [a, b]
How can we prove if x > 0, show that ln(1+x)<x (8)
-Let f(x) = ln(x)
-Then this is continuous and differentiable for all x > 0, and f’(x) = 1/x
-Applying the MVT: for any a, b > 0, such that 0<a<b, there exists c ∈ (a, b) such that (ln(b) - ln(a))/(b-a) = 1/c
-Furthermore, 1/x is decreasing for x > 0, hence for 0<a<c<b we have 1/a > 1/c > 1/b
-1/b < (ln(b) - ln(a))/(b-a) < 1/a
-Now set b = 1 + x, a = 1
-1/(1+x) < (ln(1+x))/x < 1
-Now x/(1+x) < ln(1+x) < x
How can you prove that sin(x) < x < tan(x) for x ∈ (0, π/2) (7)
-Set f(x) = sin(x), and let 0<a<b<π/2
-By the MVT, there exists c ∈ (a,b) such that (sin(b) - sin(a))/(b-a) = cos(c)
-cos(x) is decreasing in (0, π/2), so cos(b)< cos(c)<cos(a)
-Substituting, we get cos(b) < (sin(b) - sin(a))/(b-a) < cos(a)
-Set b = x, a = 0, and this becomes cos(x) < sin(x)/x < 1
-1<x/sin(x)<1/cos(x)
-sin(x)<x<tan(x)
What is the Cauchy mean value theorem (2)
-Let f, y: ℝ -> ℝ be continuous over [a, b] and differentiable over (a, b)
-Then there exists c ∈ (a, b) such that (f(b) - f(a))/(g(b) - g(a)) = f’(c)/g’(c), where g’(x) ≠ 0
What is the link between proving the MVT and CMVT (2)
-We can’t just prove the CMVT by using the MVT applied to f and g, because the value of C we get for F may be different to the value of c we get for g
-However, we can use the CMVT to prove the MVT by setting g(x) = x
What is L’Hopitals rule (2)
-If f(a), g(a) = 0, and f’(a), g’(a) both exist, then:
-limx->a f(x)/g(x) = f’(a)/g’(a)
How can we prove L’Hopitals rule (6)
-Use the definition of the derivative
-We know f(a) = g(a) = 0
-Therefore, f(x)/g(x) = (f(x) - f(a))/(g(x) - g(a))
= ((f(x) - f(a))/(x-a))/((g(x) - g(a))/(x-a))
-Then limx->a = limx->a ((f(x) - f(a))/(x-a))/((g(x) - g(a))/(x-a))
=f’(a)/g’(a)
How can we calculate limx->0 (1-x)1/x (4)
-Let y = (1-x)1/x
-ln(y) = ln(1-x)/x, this being of type “0/0”
-Applying LHR, limx->0 ln(y) = limx->0 -1/1-x = -1
-So ln(y) -> -1, and y-> e-1
What is the first mean value theorem (2)
-Say f is continuous on [a, b] and differentiable on (A, b)
-There exists c ∈ (a, b) such that: f(b) = f(A) + (b-a)f’(c)
How can we use the first MVT to create a constant approximation for f(x) near a (3)
-f(b) = f(A) + (b-a)f’(c)
-f(b) ≈ f(a) + a correction term
-Ignoring the correction term gives us a constant approximation for f(x) near a
What is the second MVT (2)
-If f and f’ are continuous in [a, b], and f’’ exists in (a, b):
-Then, there exists c ∈ (a, b) such that f(b) = f(a) + (b-a)f’(a) + (b-a)2/2f’‘(c)
What is taylor’s theorem (2)
-Suppose f is continuous and differentiable to order n-1 over [a, b], and differentiable to order n over (a, b)
-Then there exists c ∈ (a,b) such that f(b) = f(a) + (b-a)f’(a) + (b-a)2/2f’‘(c) + … + ((b-a)n/n!)fn(c)
What can we use taylors theorem for (1)
-Use it to derive degree n approximations to a function close to some fixed value x = a
