assignment 11 qa

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course mth 173

time 1543date 24/8/10

" "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):

There are nine equations and solutions in this assignment

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Self-critique rating #$&*3

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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:

Y’=-3e^x

Y’=.02/x

Y’=cos(x)/5

I can't tell how you got your answers. You don't detail how you used the rules. You need a lot more attention to detail.

confidence rating #$&* 2.5

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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):

I forgot the neg sign on the first prob. And also I forgot the power rule first for the third prob.

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Self-critique rating #$&*3

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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 solution is .5 because the der of the equation is 5/x and when the x or 10 is subsituted you have a resultant of 5/10

confidence rating #$&* 3

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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 #$&*3

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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:

E^x does not change so by subsituting the 2 for t we have a resultant of .73

confidence rating #$&* 3

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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 #$&*3

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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:

8100 because the power rule changes the t^3 to t^2 and then you multiply the 12 times 3

So we have the result

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):

ok

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Self-critique rating #$&*3

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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 what rate is position changing when t = 0, when t = `pi/2, and when t = 4?

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Your solution:

.35

.3498

.349

you should be providing the details of your reasoning

confidence rating #$&*3

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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 don’t know why I have varying results from yours I used my calc and also used the cos(x) and subsituted all values.

you were probably in degree mode

You should be sketching the circular model for these angles. For angles which are multiples of pi/4 or pi / 6 you should know the exact values of the sine and cosine, and the two-siginificant-figure approximations to these values. For other angles you should be able to make a reasonable estimate of the sine or cosine. Then if your calculator goes haywire, you need to figure out what went wrong.

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Self-critique rating #$&*3

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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:

Y=12x^2-14x+6

Y=4cos(x)+(8/x)

Y=5e^x+15x^-6

confidence rating #$&* 2.5

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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 #$&*3

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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:

3x^2(sinx)+cosx(x^3)

E^t(cost)+-sint(e^t)

1/x(z^-3)+-3Z^2(lnz)

confidence rating #$&* 2

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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):

I should have simpolififed my answers more and also I put the first two in the wrong order at then end and the last one went craxy I don’t Know oops.

You should show the details of where you went wrong, and demonstrate what you should have been thinking.

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Self-critique rating #$&*3

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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:

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

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

Y=(1/x(sin(x)-cos(x)sin(x))/sin(x)^2

confidence rating #$&* 2

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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.

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Your solution:

I did not simplify as needed. Oops

Acutally that's not a complete bad thing; I can tell from the form of your answer how you got it. However you should still show more detail.

confidence rating #$&*3

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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):

Need to study the sin and cos rules more.

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Self-critique rating #$&*3

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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:

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):

Ok I was way off used the wrong form of differentiating

Show me the details. I can't help you see it if I can't see what you're thinking.

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Self-critique rating #$&*

3

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See my notes, which all pretty much boil down to the same thing: show the details.

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