Derivative

Question 1

For a function f to be differentiable at x=a, the function has to be..

Answer 1

If f is differentiable at a point x₀, then f must also be continuous at x₀. In particular, any differentiable function must be continuous at every point in its domain.

The converse does not hold.

Question 2

A continuous function need not be differentiable. True or False?

Answer 2

True. A continuous function need not be differentiable. For example, a function with a bend, cusp, or vertical tangent may be continuous, but fails to be differentiable at the location of the anomaly.

Question 3

Example of of a function which is continous but not differentiable.

Answer 3

An ordinary cusp on the cubic curve (semicubical parabola) x³ – y² = 0, which is equivalent to the multivalued function f(x) = ± x^3/2. This relation is continuous, but is not differentiable at the cusp.

Question 4

Rolle’s theorem

Answer 4

If a real-valued function ƒ is continuous on aclosed interval [a, b], differentiable on the open interval(a, b), and ƒ(a) = ƒ(b), then there exists a c in the open interval (a, b) such that f’**(c) = 0.

Indian mathematician Bhāskara II (1114–1185) is credited with knowledge of Rolle’s theorem.[1]

Question 5

Mean Value Theorem

Answer 5

For any function that is continuous on [a, b] and differentiable on (a, b) there exists some c in the interval (a, b) such that the secant joining the endpoints of the interval [a, b] is parallel to the tangent at c.

Question 6

Proof of Mean Value Theorem

Answer 6

* Let g(x) = f(x) − rx, where r is a constant. * Since f is continuous on [a, b] and differentiable on (a, b), the same is true for g. * Choose r so that g satisfies the conditions of Rolle’s theorem. * Since g is differentiable and g(a) = g(b), there is some c in (a, b) for which g′(c) = 0, and it follows from the equality g(x) = f(x) − rx that,

Question 7

Which theorem do we use to prove the Mean Value Theorem

Answer 7

Rolle’s theorem

Question 8

If f is differentiable at x = a then

Answer 8

The limit h approaching 0, (f(a+h) - f(a))/h

Question 9

Product rule

Answer 9

(f*g)’ = f’*g + g’f

Question 10

Proof of product rule

Answer 10

1. Let h(x) = f(x) g(x),
2. suppose that f and g are each differentiable at x
3. We want to prove that h is differentiable at x and that its derivative h’(x) is given by f’(x) g(x) + f(x) g’(x).
4. Add f(x)g(x+deltax) - f(x) g(x+deltax) to the numerator to permit its factoring
5. Properties of limits are used.

Question 11

What is (f/g)’?

Answer 11

(g*f’ - f*g’)/g²

Question 12

(f(g(x))’

Answer 12

f’(g(x)) ( g’(x)

Question 13

(f^-1(x))’

Answer 13

1. Let y = f^-(x)
2. Then f^-1(x))’ = 1/f’(y)

f(g(x)

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Question 14

Why should we be more careful than usual in our statement of the differentiability result for inverse functions?

Answer 14

1. The differentiation formula for the inverse function involves division by f ‘(f -1(x)).
2. We must therefore assume that this value is not equal to zero. There is also a graphical explanation for this necessity.

Question 15

• f(x) = x^a.*
• f”(x) = ?*

Answer 15

ax^a-1

Question 16

Proof of the power rule (derivatives). This is not obvious.

Answer 16

1. Restrict n to be a positive integer.
2. Let f’ (x) = xⁿ
3. The definition of the derivative f’(a) is given by the limit form: f’(a) = limit_x->a (f(x) - f(a))(x-1)
4. Write the expansion of xⁿ - aⁿ=(x-a)(x^n-1+ …
5. Notice that there are n terms in the second factor
6. Plug this into the formula for the derivative
7. Notice that we can cancel x-a factor and
8. Then compute the limit.
9. After combining the exponents in each term we can see that we get the same term.
10. So, then recalling that there are n terms in second factor we can see that we get what we claimed it would be. f’(x) = nx^n-1

Question 17

fix) = sin(x), f’(x) = ?

Answer 17

cos (x)

Question 18

Proof of the limit of sin(theta)/theta where theta approaches 0.

Answer 18

1. Use the Squeeze Theorem.
2. Assume that theta is between 0 and Pi/2.
3. By assuming that theta is positive we’re actually going to first prove that the above limit is true if it is the right-hand limit.
4. Draw a unit circle circumscribed by an octagon with a small slice marked out as shown below.
5. The circumference of the circle is less than the length of the octagon.
6. Take a slice of the figure marked out.
7. Then the length of the portion of the circle included in the slice must be less than the length of the portion of the octagon included in the slice.
8. Denote the portion of the circle by arc AC and the lengths of the two portion of the octagon shown by |AB| and |BC|.
9. Length arc AC is less than |AB| + |BC|
10. Extend the lines AB and OC (O is the center of the circle) as shown below and call the point that they meet D. The triangle now formed by AOD is a right triangle.
11. The triangle BCD is a right triangle with hypotenuse BD and so we know |BC| < |BC| .
12. Notice that AB + BD = AD
13. arc AC < |AB| + BC < AB + BD

AD = AO * tan theta.

arc AC < AO * tan theta

AO = 1, so arc AC < tan(theta)

arc AC = AO * theta theta = arc AC < sin(theta)/cos(theta)

14. Rearranging we get cos(theta) < sin(theta)/theta < 1
15. Let’s connect A and C with a line and drop a line straight down from C until it intersects AO at a right angle and let’s call the intersection point E. This is all shown in the figure below.

Question 19

f(x) = tan(x), f’(x)=?

Answer 19

sec²(x)

Question 20

f(x) = e^x, f’(x) = ?

Answer 20

e^x

Question 21

Proof of the derivative of e^x

Answer 21

1. Use the limit definition of the derivitave; as h approaches zero, what is (e^x+h) - e^x)/h
2. Split the addition of exponents into multiplication of the same base
3. Factor out an e^x. We can put the e^x in front of the limit
4. As h approaches 0, the limit will get closer to 0/0 which is an indeterminant form
5. Plug in the point (0,1) and see the function’s behavior at that point.
6. e is the unique positive number for which (e^x-1)/x =1 in the limit of x approaching zero.
7. We can substitute 1 for the limit

Question 22

Average rate of change of f on [a,b]

Answer 22

(f(b) - f(a))/(b-a)

Question 23

If f’(a)== 0, then ..

Answer 23

There is a horizontal tangent at x=a

Question 24

f’(x) < 0 for a < x

Answer 24

f is decreasing on the interval a < x < b

Question 25

f’‘(x) > 0 for a < x < b

Answer 25

The graph of f is concave upward on the interval a < x < b

Question 26

f’‘(x) < 0 for a < x < b

Answer 26

The graph of f is concave downward on the interval a < x < b

Question 27

If

f’(a) = 0 and f’‘(a) < 0

Answer 27

f has a maximum at x - a (Second derivative test)

Question 28

f”(x₀) = 0 and f’‘(x₀) > 0

Answer 28

Suppose is a function of that is twice differentiable at a stationary point x₀ .

1. If f’‘(x₀ > 0) , then has a local minimum at x₀.
2. If f’‘(x₀ < 0) , then has a local maximum at x₀.

Question 29

cos(x)

Question 30

What is the strategy of the geometric proof that cos x is the derivative of sin x?

Answer 30

1. Use the definition of the sine function as the ratio of |opposite| to |hypotenuse| of the side lengths of a right triangle.
2. Draw a unit circle with angle theta.
3. Add a small amount Δθ to angle θ; let Q be the point on the unit circle at angle θ + Δθ.
4. The y-coordinate of Q is sin(θ + Δθ).
5. To find the rate of change of sin θ with respect to θ we need to find the rate of change of y = sin θ.
6. Draw a triangle PQR with perpendicular at Q; set PR=Δy
7. When Δθ is small, PQ≈ PQ. Δy = |PR| and segment PQ is a straight line approximation of the circular arc PQ.
8. If Δθ is small enough, segment PQ and arc PQ are practically the same, so |PQ| ≈ Δθ.
9. We’re trying to find Δy. Since we know the length of the hypotenuse PQ, all we need is the measure of angle QPR to solve for Δy = |PR|. Since Δθ is small, segment PQ is (nearly) tangent to the circle, and so angle OPQ is (nearly) a right angle.
10. We know that PR is vertical, we know that θ is the angle OP makes with the horizontal, and we can combine these facts to prove that angle RPQ and θ are (nearly) congruent angles.
11. The arc length Δθ is approximately equal to the length |PR| of the hypotenuse and angle RPQ is approximately equal to θ.
12. By the definition of the cosine function we get cos θ ≈ |PR| . But PR is just the vertical distance be­ Δθ | | tween Q and P, which is just the difference between sin(θ + Δθ) and sin θ. In other words, when Δθ is very small, cos θ ≈ (sin(θ + Δθ) − sin θ )/ Δθ
13. As Δθ approaches 0, segment QP gets closer and closer to arc QP and angle QPO gets closer and closer to a right angle, so the value of Δθ gets closer and closer to cos θ.
14. We conclude that: lim Δθ→0 Δθ sin(θ + Δθ) − sin θ = cos θ and thus that the derivative of sin θ is cos θ.
15. http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/1.-differentiation/part-a-definition-and-basic-rules/session-8-limits-of-sine-and-cosine/MIT18_01SCF10_Ses8d.pdf

Question 31

Derivative of cos x

Answer 31

- sin x

Answer 32

• D_x cos x=D_x sin(π/2−x)* (Sine of Complement equals Cosine)
• =−cos(π/2−x)* (Derivative of Sine Function and Chain Rule)

=−sinx (Cosine of Complement equals Sine)

Question 33

Derivative of tan x

Answer 33

D_{<span>x</span>}(tanx)=sec^{<span>2</span>}x

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when cos x ≠0.

Proof

From the definition of the tangent function:

• tan x=sin x/cos x

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From Derivative of Sine Function:

• D_{<span>x</span>}(sin x)=cos x

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From Derivative of Cosine Function:

• D_{<span>x</span>}(cos x)=−sin x

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Then:

D_{<span>x</span>}(tan x)=((cos(x) * cos (x) ) − (sin(x)*(−sinx)))/ cos²xQuotient Rule for Derivatives= ( cos^{<span>2 </span>}x+sin^{<span>2 </span>}x)/cos^{<span>2</span>}x=1/cos^{<span>2 </span>}x

Question 34

tan(x)=sin(x)/cos(x)

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1. Derivative of Sine Function which is cos(x)
2. Derivative of Cosine Function which is −sin(x).
3. Apply the Quotient Rule for Derivatives
4. Use the Sum of Squares of Sine and Cosine to simplify the numerator to 1.
5. The result is sec² x, as desired.

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Question 35

f(x) = e^x, f’(x) = ?

Answer 35

e^x