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- StudyBlue
- Canada
- University of Manitoba
- Mathematics
- Mathematics Math 1500
- Robert Borgersen
- Theorems, Derivatives, and Antiderivatives

Angelika S.

Antiderivative of 1/x

In |x|

If for a differentiable function f on some interval I, f '(x)<0 then the function is decreasing

Let x1 and x2 be any two numbers in the interval, x1<x2.

According to the definition of a decreasing function, we have to show that f(x1)<f(x2)

Because we are given that f '(x) <0, we know that f is differentiable on the interval and continuous on [x1,x2], so by the MVT, there must be a point c, x1<c<x2

f '(c)= f(x2)-f(x1) / x2 - x1

Since f '(c) < 0, the fraction on the right is

negative

Since x1>x2, x2-x1> or = 0, and so f(x2)-f(x1) <0. Thus f(x1)>f(x2), and so the function is decreasing

According to the definition of a decreasing function, we have to show that f(x1)<f(x2)

Because we are given that f '(x) <0, we know that f is differentiable on the interval and continuous on [x1,x2], so by the MVT, there must be a point c, x1<c<x2

f '(c)= f(x2)-f(x1) / x2 - x1

Since f '(c) < 0, the fraction on the right is

negative

Since x1>x2, x2-x1> or = 0, and so f(x2)-f(x1) <0. Thus f(x1)>f(x2), and so the function is decreasing

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Anti. Of cos x

Sin x

Antideriv. Sin x

- cos x

Antideriv. Sec^2 x

Tan x

Antideriv. Sec x tan x

Sec x

Antideriv. e^{x}

e^{x}

Antideriv. X^{n}

X^{n+1} / n+1

Antideriv. Csc x cot x

-csc x

Derivative of log_{a}x

In x / ln a

ln (e^{x}) derivative

X

Derivative cot x

-csc^{2}x

Derivative secx

sec x tan x

csc x Derivative

-csc x cot x

Derivative tan x

Sec^{2}x

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Derivative cos x

-sin x

Derivative sin x

cos x

Derivative ln x

1/x

Mean Value Theorem

Let f be a function that satisfies the following terms:

- f is continuous on [a,b]

-f is differentiable on (a,b)

There is a number c in (a,b) such that f'(c)= f(b)-f(a)/b-a

- f is continuous on [a,b]

-f is differentiable on (a,b)

There is a number c in (a,b) such that f'(c)= f(b)-f(a)/b-a

What is the Definition of the derivative

f(x)= f(x+h)-f(x) / h

If for a differentiable function f on some interval, f '(x)>0, then f is increasing.

Let x_{1} and x_{2} be any two numbers in the interval x_{1}<x_{2}. According to definition we must show that f(x_{2})>f(x_{1}).

Because we are given that f '(x)>0, we know that f is differentiable on the interval [x_{1},x_{2}] and so by the MVT there is a number c, x_{1}<c<x_{2}, such that

f '(c) = f(x_{2})-f(x_{1}) / x_{2}-x_{1}

Since f '(c) >0, the function on the right is positive. Since x_{2}>x_{1}, x_{2}-x_{1}> or = 0, and so f(x_{2})>f(x_{1}), and so the function is increasing.

Because we are given that f '(x)>0, we know that f is differentiable on the interval [x

f '(c) = f(x

Since f '(c) >0, the function on the right is positive. Since x

If for a differantiable function f on some interval (a,b), f'(x)=0, then f is constant (a,b). Prove that..

Proof..

Let x_{1} and x_{2} be any two numbers in (a,b) with x_{1} < x_{2}.

Since f is differentiable on (a,b), it must be differentiable on (x_{1},x_{2}) and continuous on [x_{1},x_{2}].

By applying the MVT to f on [x_{1},x_{2}], we get a number c such that x_{1}<c<x_{2} and

f'(c)= f(x_{2})-f(x_{1}) / x_{2}-x_{1}

But we assumed that f'(x)=0 for all x in (a,b) therefore, f(x_{2})-f(x_{1})=0 and so f(x_{2})=f(x_{1}).

Therefore for any two numbers x_{1} and x_{2} in (a,b), f(x_{1})=f(x_{2}) and the function is constant on (a,b)

Let x

Since f is differentiable on (a,b), it must be differentiable on (x

By applying the MVT to f on [x

f'(c)= f(x

But we assumed that f'(x)=0 for all x in (a,b) therefore, f(x

Therefore for any two numbers x

If for a differentiable function f an some interval (a,b) f '(x) <0, then f is decreasing

Let x1 and x2 be any two numbers in the interval, x1<x2. According to the definition of a decreasing function, we have to show that f(x1)>f(x2)

Because we are given that f '(x)< 0, we know that f is differentiable on the interval and specifically on [x1,x2] and so by the MVT there is a point c, x1<c<x2

Such that f '(x)= f(x2)-f(x1) / x2 - x1

Since f'(c)<0, the fraction on the right is negative. Since x2>x1, x2-x1> or = 0, and so f(x2)-f(x1)<0. Thus f(x1)>f(x2), and so

the function is decreasing.

Because we are given that f '(x)< 0, we know that f is differentiable on the interval and specifically on [x1,x2] and so by the MVT there is a point c, x1<c<x2

Such that f '(x)= f(x2)-f(x1) / x2 - x1

Since f'(c)<0, the fraction on the right is negative. Since x2>x1, x2-x1> or = 0, and so f(x2)-f(x1)<0. Thus f(x1)>f(x2), and so

the function is decreasing.

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