qa11

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course Mth 173

2/27 9:59 PM.

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:

y= -3e^x, exponential, the derivative is the same, so y'=-3e^x

y=.02ln(x), y'= .02/x

y= 7x^3, y'= 7 * 3 * x^2, y'= 21x^2

y=sin(x)/5, y= (1/5)sin(x), y'=(1/5)cos(x)

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: OK

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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: y' = 5/x, y'=1/2 at t=10

so it is raising y .5units/second.

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: OK

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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: y'=(1/10)e^t

y'=(1/10)e^2

y'= 7.389056099 at t=2

about .7389units/1

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: OK

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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:

y'=36t^2

y'= 36 * 225

y'= 8100 at t=15

8100units/1

confidence rating #$&*:3

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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:OK

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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:

y'= .35cos(t)

y'=.35 at t=0

.35units/1

y'= .349868475 at t= PI/2

about .3499/1

y'= .349147417 at t=4

about .349/1

???when we say cos(t), how is that evaluated. I know how to plug it into a calculator, but i want to know how to actually do it. as well as sin. those...I don't even know what to call them, functions? still give me a little trouble.

@&

The sine and cosine are indeed functions.

You cannot calculate sines and cosines, except for a few special angles.

If you draw a circle of radius 1 centered at the origin, and a half-line from the origin directed at angle theta as measured counterclockwise from the positive x axis, the point where the line intersects the circle will have coordinates (x, y), with x = cosine(theta) and y = sine(theta).

If the angle happens to be 30 degrees, which is the same as pi/6 radians, a triangle from the origin to the point of intersection (x, y), then to the point (x, 0) and back to the origin, will be half of an equilateral triangle (the equilateral triangle of which it is half will have vertices (0, 0), (x, y) and (-x, y) ). It follows that the hypotenuse of the triangle will have length 1 and the vertical leg length 1/2 (that leg being half the length of the side of the equilateral triangle). By the Pythagorean Theorem is it easy to determine that the length of the remaining leg is sqrt(3) / 2.

Thus the point (x, y) is (sqrt(3) / 2, 1/2), and cos(30 degrees) = cos(pi/6) = sqrt(3) / 2, with sin(30 degrees) = sin(pi/6) = 1/2.

Another argument involving a triangle shows that cos(45 degrees) = cos(pi/4) = sqrt(2) / 2, and sin(45 degrees) = sin(pi/4) = sqrt(2) / 2.

It is easy to show that sin(0) = 0, cos(0) = 1, sin(90 degrees) = sin(pi/2 radians) = 1, and cos(90 degrees) = cos(pi/2) = 0.

Basic sum-of-angle formulas, half-angle formulas and others allow calculation of sines and cosines of other angles, based on these known exact values.

You can easily estimate sines and cosines by drawing the unit circle and sketching a line to approximate the angle. Estimating the x and y coordinates of the intersection point provides estimates of the sine and cosine of that angle.

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I believe I provided you with some guidance for learning about sines and cosines, accompanying your Query for that section of the text. I do recommend that, one way or another, you learn a good bit of trigonometry, which is in fact prerequisite for this course and which you will use this semester, and even more extensively next semester. You don't need to know much trigonometry to be able to find the derivatives and, later, the integrals of these functions, but to interpret them and apply them you do need to know the subject.

If you can't locate my recommendations for the trigonometry let me know and I can get you another copy.

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confidence rating #$&*:2

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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):

When I plugged it into my calculator, I received different values. How would I evaluate these with pencil and paper?

@&

The angles are in radians, not degrees. You probably had your calculator in degree mode. Change it to radian mode.

cos(0) and cos(pi/2) are respectively 1 and 0, as is obvious from the circular model (and you need to understand the circular model well enough to see that this is obvious).

cos(4) can be estimated by sketching the circular model, with the line drawn at 4 radians. However to get an accurate value of cos(4), you will use your calculator. There is no reasonable way to calculate this value.

*@

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Self-critique Rating: 1

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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 and y'= 14x+6

y'=12x^2 - 14x + 6 for y= 4x^3 - 7x^2 + 6x

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y=4sin(x) + 8ln(x)

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

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

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y=5e^x - 3x^-5

y'=5e^x, y'=15x^-6

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

confidence rating #$&*: 3

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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:OK

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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:

y= x^3 * sin(x)

y'=3x^2, y' =cos(x)

y'= 2x^2 * sin(x) + cos(x) * x^3

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y= e^t * cos(t)

y'=e^t, y'=-sin(t)

y'= e^t * cos(t) - sin(t) * e^t

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y=ln(z) * z^-3

y'= 1/z, y'= -3z^-4

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

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 did not simplify, I didn't see what could be simplified in this format, I should write it down on paper when it gets a little more complex like it is.

@&

Paper is always the gold standard. We can see simple expressions in typewriter format, but at a certain level of complexity we absolutely need to write the expressions out.

*@

@&

Of course going back and forth between paper and typewriter notations makes us better at both.

*@

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Self-critique Rating: 2

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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

y'= e^t, y'= 5t^4

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

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y=sin(x)/cos(x)

y'= cos(x), y'=-sin(x)

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

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

---this part looks familiar to me, but I don't remember exactly what, so I'm going to look it up in my book. its a^2 + b^2 = c^2 where cos^2 and sin^2 = 1

1/cos(x)^2

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y= ln(x)/ sin(x)

y'=1/x, y'=cos(x)

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

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

???is there something else that can be done, I feel like there is more to it

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.

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Your solution:

im not sure how you simplified the first one

@&

It's not clear from the information I see here, in your solution and in your question, what step(s) you do and do not understand.

You're welcome to submit a question form with a little more detail regarding this simplification.

*@

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):OK

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Self-critique Rating: OK

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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:

y=4ln(x)/sin(x) - sin(x) * cos(x)

y'=4/x, y'=cos(x) y'= cos(x), y'= -sin(x)

y'=(4/x * sin(x) - cos(x) * 4ln(x) ) / sin(x)^2

y' = (4sin(x)/x - cos(x) * 4ln(x) ) / sin(x)^2

cos(x) * cos(x) - sin(x) * sin(x)

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

y'= ((4sin(x)/x - cos(x) * 4ln(x) ) / sin(x)^2) - cos(x)^2 - sin(x)^2

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y= 3e^t/ t+ 6ln(t)

???I am unsure how this one would be solved, the t by itself also is throwing me off a bit.

i guess since the derivative stays the same, ill stick it with

y'= 3e^t/t, y'= 6/t

@&

e^t / t is a quotient, with e^t as numerator and t as denominator. So you will apply the quotient rule to this term.

The derivative of ln(t) is done separately, since ln(t) is a separate term.

*@

y'= (3e^t + 6)/t

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y=-5t^5/ln(t) + sin(t) / 5

y'-25t^4, y'= 1/t, y'=cos(t)

????still not sure what to do with the individual 5, just like with the t

@&

For t^5 / ln(t) you apply the quotient rule. Of course you'll multiply the result by -5.

sin(t) / 5 is just 1/5 * sin(t). So the derivative is 1/5 the derivaitve of sin(t).

*@

((-25t^4 * ln(t) - 1/t * -5t^5)/ln(t)^2) + cos(t)/5

confidence rating #$&*:1

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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):

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Self-critique Rating:"

@&

As you see, the process of finding derivatives gives you a bit of a workout with your algebra. For most students, this is the course where algebra skills finally get well-honed.

Be sure you recognize and apply the product and quotient rules very carefully and systematically. If one function is multiplied by another, you use the product rule. If one function is divided by another, you use the quotient rule.

See also my note regarding the trigonometry.

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