QA_011

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course mth 173

7/19 1

011. Rules for calculating derivatives of some functions.

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Question: `qNote that there are 9 questions in this assignment.

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Self-critique (if necessary):

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Self-critique Rating:

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Question: `q001. The most basic functions you studied precalculus were:

the power functions y = x^n for various values of n,

the exponential function y = e^x,

the natural logarithm function y = ln(x), and

the sine and cosine functions y = sin(x) and y = cos(x).

We have fairly simple rules for finding the derivative functions y ' corresponding to each of these functions. Those rules are as follows:

If y = x^n for any n except 0, then y ' = n x^(n-1).

If y = e^x then y ' = e^x (that's right, the rate of change of this basic exponential function is identical to the value of the function).

If y = ln(x) then y ' = 1/x. If y = sin(x) then y ' = cos(x).

If y = cos(x) then y ' = - sin(x).

There are also some rules for calculating the derivatives of combined functions like the product function x^5 * sin(x), the quotient function e^x / cos(x), or the composite function sin ( x^5).

We will see these rules later, but for the present we will mention one easy rule, that if we multiply one of these functions by some constant number the derivative function will be the derivative of that function multiply by the same constant number.

Thus for example,

since the derivative of sin(x) is cos(x), the derivative of 5 sin(x) is 5 cos(x); or

since the derivative of ln(x) is 1 / x, the derivative of -4 ln(x) is -4 (1/x) = -4 / x.

Using these rules, find the derivatives of the functions y = -3 e^x, y = .02 ln(x), y = 7 x^3, y = sin(x) / 5.

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Your solution:

The derivative rule for y = e^x is the identical function, therefore the derivative for y = -3e^x would be y’ = -3e^x.

The derivative rule for y = ln(x) is y’ = 1/x then the derivative for the function y = .02ln(x) is y’ = .02(1/x) or y’ = .02/x.

The derivative rule for y = x^n is y’ = nx^(n-1), therefore the derivative for the function for y = 7x^3 is y’ = 7(3x^3-1) which is y’ = 21x^2

The derivative rule for y = sin(x) is y’ = cos(x), therefore the derivative for the function y = sin(x)/5 is y’ = cos(x)/5

confidence rating #$&*:

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Given Solution:

`aThe derivative of y = -3 e^x is -3 times the derivative of y = e^x.

Since by the given rules the derivative of e^x is e^x, the derivative of y = - 3 e^x is y ' = - 3 e^x.

The derivative of y = .02 ln(x) is .02 times the derivative of y = ln(x).

Since the derivative of ln(x) is 1 / x, the derivative of y = .02 ln(x) is y ' = .02 * 1 / x = .02 / x.

The derivative of y = 7 x^3 is 7 times the derivative of x^3. Since the derivative of x^n is n x^(n-1), the derivative of x^3 is 3 x^(3-1), or 3 x^3.

The derivative of y = 7 x^3 is therefore y ' = 7 ( 3 x^2) = 21 x^2.

The derivative of y = sin(x) / 5 is 1/5 the derivative of sin(x). The derivative of sin(x), according to the rules given above, is cos(x). Thus the derivative of y = sin(x) / 5 is y ' = 1/5 cos(x), or y ' = cos(x) / 5.

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Self-critique (if necessary):

OK.

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Self-critique Rating:

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Question: `q002. If a container is shaped so that when a certain constant water stream flows into the container, the depth function is y = 5 * ln(t), then at what rate is water rising in the container when t = 10?

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Your solution:

The derivative rule for y = ln(x) is y’ = 1/x, therefore, for this function, the derivative would read y’ = 5(1/10) which gives you a rate for the water rising in the container as .5 cm/s (unsure about units)

confidence rating #$&*:

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Given Solution:

`aThe time rate at which water is rising is the derivative y ' = dy / dt of the depth function y. Since the derivative of ln(t) is 1 / t, we have

rate = y ' = 5 * 1 / t = 5 / t.

Since the rate is y ' = 5 / t, when t = 10 the water is rising at rate y ' = 5 / 10 = .5.

If y is depth in cm and t is clock time in seconds, then the rate is y ' = dy / dt = .5 cm/sec.

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Self-critique (if necessary):

OK.

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Self-critique Rating:

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Question: `q003. If a container is shaped so that when a certain constant water stream flows into the container, the depth function is y = e^t / 10, then at what rate is water rising in the container when t = 2?

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Your solution:

Since the derivative rule for y = e^x is identical to the original function (y’ = e^x), then the derivative would look like y’ = e^(2)/10 which would give you a rate for approximately .7389 cm/s.

confidence rating #$&*:

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Given Solution:

`aThe time rate at which water is rising is the derivative y ' = dy / dt of the depth function y. Since the derivative of e^t is e^t, we have

rate = y ' = e^t / 10.

Since the rate is y ' = e^t / 10, when t = 2 the water is rising at rate y ' = e^2 / 10 = .73, approx.

If y is depth in cm and t is clock time in seconds, then the rate is y ' = dy / dt = .73 cm/sec.

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Self-critique (if necessary):

OK.

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Self-critique Rating:

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Question: `q004. If the altitude of a certain rocket is given as a function of clock time t by y = 12 * t^3, then what function gives the rate of altitude change, and at what rate is the altitude changing when t = 15?

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Your solution:

For this problem you start out by finding the derivative for the function y = t^3 which is y’ = 3t^2. After that, you take the remaining part of the original function, which is 12, and multiply it by the derivative you just found, which then gives you a result of y’ = 36t^2. This is the function that gives the rate of altitude. When the t = 15 is put in for the value of t in the rate function (y’ = 36(15)^2) you get a rate of 8,100 m/s. (unsure about units)

confidence rating #$&*:

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Given Solution:

`aThe time rate at which altitude is changing is the derivative y ' = dy / dt of the depth function y. Since the derivative of t^3 is 3 t^2, we have

rate = y ' = 12 * (3 t^2) = 36 t^2.

Since the rate is y ' = 36 t^2, when t = 15 the altitude is changing at rate y ' = 36 * 15^2 = 8100, approx.

If y is altitude in feet and t is clock time in seconds, then the rate is y ' = dy / dt = 8100 ft/sec.

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Self-critique (if necessary):

Used incorrect units, but wasn’t sure if it mattered.

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Self-critique Rating:

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Question: `q005. If the position of a certain pendulum is given relative to its equilibrium position by the function y = .35 sin(t), then what function gives the corresponding rate of position change, and at what rate is position changing when t = 0, when t = `pi/2, and when t = 4?

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Your solution:

Since the derivative rule for y = sin(x) is y’ = cos(x), then the derivative for this function would be y’ = .35cos(x). The rate at which the position is changing at t = 0 is represented by y’ = .35cos(0) which is a rate of .35 cm/s. The rate at which the position is changing at t = π/2 is represented by y’ = .35cos (π/2), which is a rate of approximately 0 cm/s. The rate at which position is changing at t = 4 is represented by y’ = .35cos (4) which is a rate of -.228 cm/s. (unsure about units)

confidence rating #$&*:

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Given Solution:

`aThe time rate at which position is changing is the derivative y ' = dy / dt of the position function y. Since the derivative of sin(t) is cos(t), we have

rate = y ' = .35 cos(t).

Since the rate is y ' = .35 cos(t),

When t = 0 the position is changing at rate y ' = .35 cos(0) = .35.

When t = `pi/2 the position is changing at rate y ' = .35 cos(`pi/2) = 0.

When t = 4 the position is changing at rate y ' = .35 cos(4) = -.23.

If y is position in cm and t is clock time in seconds, then the rates are .35 cm/s (motion in the positive direction), -.35 cm/s (motion in the negative direction), and -.23 cm/s (motion in the negative direction but not quite as fast).

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Self-critique (if necessary):

I got all of the correct values, but I’m a little unsure about a part of the last statement. Where did the -.35 (motion in the negative direction) come from? Was this the representative rate of motion for the rate of 0?

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Self-critique Rating:

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There's an apparent editing error in the given solution. It wasn't asked, but if t = pi then y ' = .35 * cos(pi) = -.35.

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That's where the -.35 came from, but it shouldn't have appeared in the given solution.

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Question: `q006. Another rule is not too surprising: The derivative of the sum of two functions is the sum of the derivatives of these functions. What are the derivatives of the functions y = 4 x^3 - 7 x^2 + 6 x, y = 4 sin(x) + 8 ln(x), and y = 5 e^x - 3 x^-5?

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Your solution:

Since y = 4x^3 - 7x^2 + 6x is just the sum of functions, the derivative is going to be the sum of these functions. The derivatives of each part are as follows: 12x^2, 14x, and 6. Then, you simply combine all of the derivatives, y’ = 12x^2 - 14x + 6.

When each part of the function y = 4sin(x) + 8ln(x) are divided up into the derivatives of 4 cos(x) and 8/x, you can simply combine them back together to get your overall derivative of y’ = 4 cos (x) + 8/x.

The function of y = 5e^x - 3x^(-5) can be put into the individual derivatives of 5e^x and -15x^(-6) then combined to form the overall derivative of y’ = 5e^x + 15x^(-6).

confidence rating #$&*:

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Given Solution:

`aSince y = 4 x^3 - 7 x^2 + 6 x is the sum of the functions 4 x^3, -7 x^2 and 6x, whose derivatives are 12 x^2, -14 x and 6, respectively, we see that y ' is the sum of these derivatives:

y ' = 12 x^2 - 14 x + 6.

Since y = 4 sin(x) + 8 ln(x) is the sum of the functions 4 sin(x) and 8 ln(x), whose derivatives are , respectively, 4 cos(x) and 8 / x, we see that y ' is the sum of these derivatives:

y ' = 4 cos(x) + 8 / x

Since y = 5 e^x - 3 x^-5 is the sum of the functions 5 e^x and 3 x^-5, whose derivatives are, respectively, 5 e^x and -15 x^-6, we see that y ' is the sum of these derivatives:

y ' = 5 e^x + 15 x^-6.

Note that the derivative of x^-4, where n = -4, is n x^(n-1) = -4 x^(-4-1) = -4 x^-5.

STUDENT QUESTION

I understand how to take the derivative but I am confused by what is meant by the statement of “The derivative of the sum

of two functions is the sum of the derivatives of these functions”

What does this mean?

INSTRUCTOR RESPONSE

You used this property without really thinking about it when you found the derivative of 4 sin(x) + 8 ln(x).

4 sin(x) + 8 ln(x) is the sum of two functions, 4 sin(x) and 8 ln(x).

Each function has a derivative:

• (4 sin(x)) ' = 4 cos(x) and

• (8 ln(x)) ' = 8 * 1/x = 8/x.

The stated property assures you that the derivative of the sum 4 sin(x) + 8 ln(x) is the sum 4 cos(x) + 8 / x of the two derivatives.

You also used the property in each of the other solutions, again really with even noticing that you had used it. It's pretty automatic.

However the rules for product and quotient functions are not what you would at first expect them to be, as you will see in the next question.

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Self-critique (if necessary):

OK.

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Self-critique Rating:

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Question: `q007. The rule for the product of two functions is a bit surprising: The derivative of the product f * g of two functions is f ' * g + g ' * f. What are the derivatives of the functions y = x^3 * sin(x), y = e^t cos(t), and y = ln(z) * z^-3?

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Your solution:

The derivatives for the individual parts of the function y = x^3 * sin (x) are 3x^2 and cos (x), but since you must follow the rule of finding the derivative for the product of two functions, you must follow the formula f’ * g + g’ * f.

F’ = 3x^2

G = sin(x)

G’ = cos(x)

F’ = x^3

Y’ = (3x^2)(sin (x)) + (cos (x))(x^3)

For the function of y = e^t * cos (t), the values for each part are as follows:

F’ = e^t

G = cos(t)

F = 0

F = e^t, not 0

G’ = sin (t)

Y’ = (e^t)(cos (t)) + sin (t)

Your second term G ' F should be sin(t) e^t.

For the function y = ln(z) * z^(-3), the values for each part are as follows:

F’ = 1/z

G = ln(z)

If F ' = 1 / z, then this is F, not G.

G’ = (-3)z^(-4)

F = z^(-3)

If G ' = -3 z^-4 then this is G, not F.

Y’ = (1/z)(ln (z)) + ((-3)z^(-4))(z^(-3)

confidence rating #$&*:

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Given Solution:

`aThe derivative of y = x^3 * sin(x), which is of form f * g if f = x^3 and g = sin(x), is

f ' g + g ' f = (x^3) ' sin(x) + x^3 (sin(x)) '

= 3x^2 sin(x) + x^3 cos(x).

The derivative of y = e^t cos(t), which is of form f * g if f = e^t and g = cos(t), is

f ' g + g ' f = (e^t) ' cos(t) + e^t (cos(t) ) '

= e^t cos(t) + e^t (-sin(t)) = e^t [ cos(t) - sin(t) ].

The derivative of y = ln(z) * z^-3, which is of form f * g if f = ln(z) and g = z^-3, is

f ' g + g ' f = (ln(z)) ' z^-3 +ln(z) ( z^-3) '

= 1/z * z^-3 + ln(z) * (-3 z^-4) =

z^-4 - 3 ln(z) * z^-4 = z^-4 (1 - 3 ln(z)).

STUDENT COMMENT

I should have simplified my answers more.

INSTRUCTOR RESPONSE

Typically students don't really learn their algebra until they have to put their derivatives into standard form so they can compare their answers with the answers in the back of the book.

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Self-critique (if necessary):

#######In the second and third parts of this question I had all of the correct components in my answers, they were just rearranged and simplified more. Am I continually doing something in the incorrect way?######

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Self-critique Rating:

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You do have some errors, but you're on the right track so I don't think you'll have a lot of trouble getting this straight.

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Question: `q008. The rule for the quotient of two functions is perhaps even more surprising:

The derivative of the quotient f / g of two functions is [ f ' g - g ' f ] / g^2.

What are the derivatives of the functions y = e^t / t^5, y = sin(x) / cos(x) and y = ln(x) / sin(x)?

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Your solution:

The different parts of the function y = e^t / t^5 are:

F: e^t

F’: e^t

G: t^5

G’: 5t^4

Y’ = (e^t)(t^5) - (5t^4)(e^t)

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You've got the numerator but you appear to have left off the denominator.

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The different parts of the function y = sin(x) / cos(x) are:

F: sin (x)

F’: cos (x)

G: cos (x)

G’: sin (x)

Y’ = (cos (x))(cos (x)) - (sin (x))(sin (x))

Y’ = cos (x)^2 - sin (x)^2

The different parts of the function y = ln(x) / sin(x) are:

F: ln (x)

F’: 1/x

G: sin (x)

G’: cos (x)

Y’ = (1/x)(sin (x)) - (sin (x))(ln (x))

confidence rating #$&*:

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Given Solution:

`aThe derivative of y = e^t / t^5, which is of form f / g if f = e^t and g = t^5, is

(f ' g - g ' f) / g^2= ( (e^t) ' t^5 - e^t (t^5) ' ) / (t^5)^2

= (e^t * t^5 - e^t * 5 t^4) / (t^5)^2

= t^4 * e^t ( t - 5) / t^10

= e^t (t-5) / t^6..

The derivative of y = sin(x) / cos(x), which is of form f / g if f = sin(x) and g = cos(x), is

(f ' g + g ' f) / g^2 =( (sin(x)) ' cos(x) - sin(x) (cos(x)) ' ) / (cos(x))^2 '

= (cos(x) * cos(x) - sin(x) * -sin(x) ) / (cos(x))^2 =

( (cos(x))^2 + (sin(x))^2 ) / (cos(x))^2 =

1 / cos(x)^2.

Note that we have used the Pythagorean identity (sin(x))^2 + (cos(x))^2 = 1.

The derivative of y = ln(x) / sin(x), which is of form f / g if f = ln(x) and g = sin(x), is

(f ' g + g ' f) / g^2 =( (ln(x)) ' sin(x) - ln(x) (sin(x)) ' ) / (sin(x))^2 =

(sin(x) * 1/x - ln(x) * cos(x) ) / (sin(x))^2 =

( sin(x) / x - ln(x) cos(x) ) / (sin(x))^2 =

1 / ( x sin(x)) - ln(x) cos(x) / (sin(x))^2. Further simplification using the tangent function is possible, but the answer here will be left in terms of the sine and cosine functions.

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Self-critique (if necessary):

######The first part of this question is proving very confusing to me, it seems as if my answer isn’t even close to the given solution#######

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Self-critique Rating:

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Your only error is that you've left off the denominator g^2.

(e^t)(t^5) - (5t^4)(e^t)

should be

[ (e^t)(t^5) - (5t^4)(e^t) ] / (t^5)^2

and

cos (x)^2 - sin (x)^2

should be

[ cos (x)^2 - sin (x)^2 ] / (cos(x))^2.

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Of course some simplification will follow.

In any case you're on the right track.

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Question: `q009. Combining the above rules find the derivatives of the following functions: y =4 ln(x) / sin(x) - sin(x) * cos(x); y = 3 e^t / t + 6 ln(t), y = -5 t^5 / ln(t) + sin(t) / 5.

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Your solution:

The derivative of the product rule side is

Y’ = (4/x)(sin (x)) - (cos (x))(4ln(x))

And the derivative of the quotient side is

Y’ = cos (x)^2 + sin (x)^2

When combined, they form an overall derivative of:

Y’ = 4/ (x sin (x)) - (cos (x)) (4 ln (x)) / cos (x)^2 + sin (x)^2

@&

4 ln(x) / sin(x) is a quotient, not a product.

sin(x) * cos(x) is a product, not a quotient.

You seem to have reversed these terms.

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The derivative of the quotient side of y = 3e^t / t + 6ln (t) is:

Y’ = (3e^t)(t-1) - (3e^t)/(t^2)

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Should be

Y’ =( (3e^t)(t) - (3e^t) ) / (t^2)

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The derivative of the remaining portion is:

6/t

Y’ = (3e^t)(t-1) - (3e^t)/(t^2) + 6/t

The derivative of the quotient side of y = -5t^5 / ln(t) + sin(t)/5 is:

Y’ = (-25t^4)(ln(t)) - (1/t)(-5t^5) / (ln(t)) ^ 2

The derivative of the remaining portion is:

Y’ = cos(t)/5

Therefore the derivative of the entire function is:

Y’ = (-25t^4)(ln(t)) - (1/t)(-5t^5) / ln(t)^2 + cos(t)/5 or simplified y’ = (-25t^4)(ln(t)) + 5t^4 / ln(t)^2 + cos(t)/5

confidence rating #$&*:

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Given Solution:

`aSince the derivative of sin(x) / ln(x) is 1 / ( x sin(x)) + ln(x) cos(x) / (sin(x))^2 ), as just seen, and the derivative of sin(x) * cos(x) is easily seen by the product rule to be -(sin(x))^2 + (cos(x))^2, we see that the derivative of y = 4 sin(x) / ln(x) - sin(x) * cos(x) is y ' = 4 [ 1 / ( x sin(x) - ln(x) cos(x) / (sin(x))^2 ] - ( -(sin(x))^2 + (cos(x))^2 ) =

4 / ( x sin(x) ) - 4 ln(x) cos(x) / ( sin(x))^2 + (sin(x))^2 - (cos(x))^2.

Further rearrangement is possible but will not be done here.

The derivative of 3 e^t / t is found by the quotient rule to be ( 3 e^t * t - 3 e^t * 1 ) / t^2 = 3 e^t ( t - 1) / t^2, the derivative of 6 ln(t) is 6 / t, so the derivative of y = 3 e^t / t + 6 ln(t) is therefore

y ' = 3 e^t ( t - 1) / t^2 + 6 / t.

Since the derivative of -5 t^5 / ln(t) is found by the quotient rule to be ( -25 t^4 ln(t) - (-5 t^5 ) * ( 1 / t ) ) / (ln(t))^2, and the derivative of sin(t) / 5 is cos(t) / 5, we see that the derivative of y = -5 t^5 / ln(t) + sin(t) / 5 is

y ' = (-25 t^4 ln(t) - (-5 t^5 ) * ( 1 / t ) ) / (ln(t))^2 + cos(t) / 5 =

-25 t^4 ln(t) + 5 t^4 / (ln(t))^2 + cos(t) / 5.

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Self-critique (if necessary):

In the first part of the given solution, I’m not exactly sure where the second “sin (x)^2” came from.

In the second part of the given solution, I have 1 more 3e^t than the given solution and I’m not sure how it cancels out. If it says 3e^t - 3e^t, if they’re going to cancel out in any way, then shouldn’t both of them disappear?

I felt pretty comfortable with the final part of this question.

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Self-critique Rating:

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Self-critique (if necessary):

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Self-critique rating:

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Self-critique (if necessary):

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Self-critique rating:

#*&!

@&

You're making some mistakes but you're very much on the right track. Just a little more practice, which you'll get from your text, and you'll be in good shape on these rules.

Check my notes.

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