Calculus Computing Derivatives.

[andylpsx]

I have this assignment that is try until you pass. The problems change all the time and this is only practice. I am not feeling super confident in my ability to solve all of the problems I am supposed to solve. I will post each question (10 questions, and my answers, I will also put if they are right or wrong. This is online submitting so simplification isn't supposed to matter. It just has to be an equivalent answer. So if it is x^-2 I can leave it as that instead of 1/(x^2). Anyways onto the assignment.

Answers in italics and questions in Bold.

1) (e^(-6u)-9)cos(7u)
(-6e^(-6u))(cos(7u))+(e^(-6u)-9)(-7sin(7u))
I used the product rule and chain rule
Got correct

2) ((4t^2-6t)+5)(3t^6+2)
((8t-6)(3t^6+2)-((4t^2-6t)+5)(18t^5))/(3t^6+2)^2
I used the quotient rule.
Got correct

3)sqrt(cos(7y)) OR (cos(7y))^.5
(-7sin(7y))/sqrt(.5cos(7y)) OR (-7sin(7y))/(.5cos(7y))^.5
Chain rule

4)(w^6)/(e^(w)sin(w))
((6w^5)(e^(w)sin(w))-(( w^6)(e^(w)cos(w))+(e^(w)sin(w))))/(e^(w)sin(w))^2
Quotient rule and Chain rule (sorry for the excessive amount of parenthesis, I am copying the answers I am typing into the computer and for some reason our software wasn't picking up terms correctly unless I had it like this.)

5) e^(-2x)cos(x^5)
(-2e^(-2x))(cos(x^5))+(e^(-2x))(-5x^4sin(x^5))
Product rule and Chain Rule
Got correct

6) ln(v^2 e^(-6v) -3cos(v))
((2v)(e^(-6v))(-3cos(v))+(v^2)(-6e^(-6v))(-3cos(v))+(v^2)(e^(-6v))(3sin(v)))/(v^2e^(-6v)-3cos(v))
chain rule and product rule

7) [sin(e^(-7y)-5)]^3
3(sin(e^(-7y)-5))^2(cos(e^(-7y)-5))(-7e^(-7y))
double chain rule
Got correct

8) ((cos(v))^6)/(6v^4-2v)
(((6(cos(v))^5)-sin(v))(6v^4-2v)-((cos(v))^6)(24v^3-2))/(6v^4-2v)^2
chain rule and quotient rule

9) Find Dy/Dx for (cos(3x)+sin(y))+x^6y^3=3
(3sin(x)-((6x^5)(y^3)))/((x^5)(3y^3)(-cos(y)))
Chain Rule and Algebra

10) ((3e^(2y)+5arctan(y))+4)-4y^.2
(3sin(x)-((6x^5)(y^3)))/((x^5)(3y^3)(-cos(y)))
not really sure what I used, just took the derivative of each term.


Besides questions 1,2,5 and 7 I missed all the rest. I am not looking for the answers directly but rather how to get the answer and if possible if you could show me my mistake. I don't really have work for these because I just did them all in 1 step on scratch paper. Also I used no calculators though there was no need to use one, but they are not supposed to be used. I am mostly curious on what my mistakes where. I can see the right answers but they do me no good because it doesn't show how to get to the right answer. Can I get any help. I know it's asking a lot.

 

[stapel]

What were the questions? I can see a list of expressions, but what were the instructions for each? What are you supposed to be doing with these expressions? (Differentiation would be of a function, such as "y=(something)", which is not what has been posted.)

 

[andylpsx]

What were the questions? I can see a list of expressions, but what were the instructions for each? What are you supposed to be doing with these expressions? (Differentiation would be of a function, such as "y=(something)", which is not what has been posted.)

All of them but the one that says find Dy/Dx are "Find the Derivative."

 

[stapel]

3)sqrt(cos(7y)) OR (cos(7y))^.5
(-7sin(7y))/sqrt(.5cos(7y)) OR (-7sin(7y))/(.5cos(7y))^.5

How are you getting a 1/2 inside the radical...?

4)(w^6)/(e^(w)sin(w))
((6w^5)(e^(w)sin(w))-(( w^6)(e^(w)cos(w))+(e^(w)sin(w))))/(e^(w)sin(w))^2

I think the second line above means the following:

. . . . .\(\displaystyle f(w)\, =\, \dfrac{w^6}{e^w \sin(w)}\)

. . . . .\(\displaystyle f'(w)\, =\, \dfrac{(6w^5)(e^w \sin(w))\, -\, (w^6(e^w \cos(w)\, +\, e^w \sin(w))}{(e^w \sin(w))^2}\)

If so, then I think there's something you can factor out and cancel off. You should also be able to multiply things out and combine like terms.

6) ln(v^2 e^(-6v) -3cos(v))
((2v)(e^(-6v))(-3cos(v))+(v^2)(-6e^(-6v))(-3cos(v))+(v^2)(e^(-6v))(3sin(v)))/(v^2e^(-6v)-3cos(v))

I think you've typed this:

. . . . .\(\displaystyle f(v)\, =\, \ln\left(v^2 e^{-6v}\, -\, 3\cos(v)\right)\)

. . . . .\(\displaystyle f'(v)\, =\, \dfrac{(2v)(e^{-6v})(-3\cos(v))\, +\, (v^2)(-6e^{-6v})(-3\cos(v))\, +\, (v^2)(e^{-6v})(3\sin(v))}{v^2 e^{-6v}\, -\, 3\cos(v)}\)

I'm not clear on how the subtraction in the original function turned into a multiplication in the derivative...?

8) ((cos(v))^6)/(6v^4-2v)
(((6(cos(v))^5)-sin(v))(6v^4-2v)-((cos(v))^6)(24v^3-2))/(6v^4-2v)^2

I think the intended statements are as follows:

. . . . .\(\displaystyle f(v)\, =\, \dfrac{\cos^6(v)}{6v^4\, -\, 2v}\)

. . . . .\(\displaystyle f'(v)\, =\, \dfrac{\left(6 \cos^5(v)\, -\, \sin(v)\right)(6v^4\, -\, 2v)\, -\, \cos^6(v)(24v^3\, -\, 2)}{(6v^4\, -\, 2v)^2}\)

I think you've converted what should be multiplication by a negative sine into subtraction. Check your math in the numerator.

9) Find Dy/Dx for (cos(3x)+sin(y))+x^6y^3=3
(3sin(x)-((6x^5)(y^3)))/((x^5)(3y^3)(-cos(y)))

You started with an "equals". You should have ended with an "equals". What did you do with the right-hand side of the equation? Where do you solve for "dy/dx"?

. . . . .\(\displaystyle (\cos(3x)\, +\, \sin(y))\, +\, x^6 y^3\, =\, 3\)

. . . . .\(\displaystyle (-\sin(3x))(3)\, +\, (\cos(y))\left(\frac{dy}{dx}\right)\, +\, 6x^5 y^3\, +\, x^6 (3y^2)\left(\frac{dy}{dx}\right)\, =\, 0\)

...and so forth.

10) ((3e^(2y)+5arctan(y))+4)-4y^.2
(3sin(x)-((6x^5)(y^3)))/((x^5)(3y^3)(-cos(y)))

I'm sorry, but I have no idea how you arrived at your answer here. Please reply showing your steps. Thank you!

 

[andylpsx]

How are you getting a 1/2 inside the radical...?


I think the second line above means the following:

. . . . .\(\displaystyle f(w)\, =\, \dfrac{w^6}{e^w \sin(w)}\)

. . . . .\(\displaystyle f'(w)\, =\, \dfrac{(6w^5)(e^w \sin(w))\, -\, (w^6(e^w \cos(w)\, +\, e^w \sin(w))}{(e^w \sin(w))^2}\)

If so, then I think there's something you can factor out and cancel off. You should also be able to multiply things out and combine like terms.


I think you've typed this:

. . . . .\(\displaystyle f(v)\, =\, \ln\left(v^2 e^{-6v}\, -\, 3\cos(v)\right)\)

. . . . .\(\displaystyle f'(v)\, =\, \dfrac{(2v)(e^{-6v})(-3\cos(v))\, +\, (v^2)(-6e^{-6v})(-3\cos(v))\, +\, (v^2)(e^{-6v})(3\sin(v))}{v^2 e^{-6v}\, -\, 3\cos(v)}\)

I'm not clear on how the subtraction in the original function turned into a multiplication in the derivative...?


I think the intended statements are as follows:

. . . . .\(\displaystyle f(v)\, =\, \dfrac{\cos^6(v)}{6v^4\, -\, 2v}\)

. . . . .\(\displaystyle f'(v)\, =\, \dfrac{\left(6 \cos^5(v)\, -\, \sin(v)\right)(6v^4\, -\, 2v)\, -\, \cos^6(v)(24v^3\, -\, 2)}{(6v^4\, -\, 2v)^2}\)

I think you've converted what should be multiplication by a negative sine into subtraction. Check your math in the numerator.


You started with an "equals". You should have ended with an "equals". What did you do with the right-hand side of the equation? Where do you solve for "dy/dx"?

. . . . .\(\displaystyle (\cos(3x)\, +\, \sin(y))\, +\, x^6 y^3\, =\, 3\)

. . . . .\(\displaystyle (-\sin(3x))(3)\, +\, (\cos(y))\left(\frac{dy}{dx}\right)\, +\, 6x^5 y^3\, +\, x^6 (3y^2)\left(\frac{dy}{dx}\right)\, =\, 0\)

...and so forth.


I'm sorry, but I have no idea how you arrived at your answer here. Please reply showing your steps. Thank you!

All of your comments make good sense, #3 I forgot that when you take 1/2 to the denominator it's 2 times a quantity. #4 should be right according to the answer, this program is no super accurate and the instructors can over rule answers like these. #9 I forgot the 3x in the sin(3x) when I typed it in. #10 which you were confused on was just me not know what arctan was and not wanting to look it up because I was trying to practice and go off of what I had in my head. so I said sec^2 instead of 1/(1+y^2).

Additionally, what do you use to show your expressions, I was trying to type them that way so it would be easier to read through.

 

[stapel]

The formatting is done with LaTeX. You can find various lessons and articles online, such as this one]/b]. On this forum, you would surround the LaTeX coding with "[tex]" tags, so writing a square root would by typed like this:

[tex]\sqrt{25}\, =\, 5[/tex]

...and would display like this:

. . . . .\(\displaystyle \sqrt{25}\, =\, 5\)

The "\," characters add blank space between the characters, so things are easier to read.

 

[Ishuda]

I went through several and its looks to me like you are getting your grouping wrong. For example, I'll work through two of them to show your mistake.

3)sqrt(cos(7y)) OR (cos(7y))^.5
(-7sin(7y))/sqrt(.5cos(7y)) OR (-7sin(7y))/(.5cos(7y))^.5
Chain rule
( -7 sin(7y) ) / sqrt(.5cos(7y) ) OR (-7sin(7y))/(.5cos(7y))^.5
The .5 is in the wrong place. The derivative of \(\displaystyle (\space f(x)\space )^{\frac{1}{2}}\) is
\(\displaystyle \frac{1}{2}(\space f(x)\space )^{-\frac{1}{2}}=0.5\space (\space f(x)\space )^{-\frac{1}{2}}\)
NOT
\(\displaystyle (\space 0.5\space f(x)\space )^{-\frac{1}{2}}\)

4)(w^6)/(e^(w)sin(w))
((6w^5)(e^(w)sin(w))-(( w^6)(e^(w)cos(w))+(e^(w)sin(w))))/(e^(w)sin(w))^2
Quotient rule and Chain rule (sorry for the excessive amount of parenthesis, I am copying the answers I am typing into the computer and for some reason our software wasn't picking up terms correctly unless I had it like this.)
( (6w^5) ( e^(w) sin(w) ) - ( ( w^6 ) ( e^(w) cos(w) ) + ( e^(w) sin(w) ) ) ) / ( e^(w) sin(w) )^2
Remove the red close and open paren to get
\(\displaystyle \frac{6\space w^5\space e^w\space sin(w)\space -\space \{\space w^6\space [\space e^w\space cos(w)\space +\space e^w\space sin(w)\space ]\space \}
}{(\space e^w\space sin(w)\space )^2}\)

 

[andylpsx]

I went through several and its looks to me like you are getting your grouping wrong. For example, I'll work through two of them to show your mistake.

3)sqrt(cos(7y)) OR (cos(7y))^.5
(-7sin(7y))/sqrt(.5cos(7y)) OR (-7sin(7y))/(.5cos(7y))^.5
Chain rule
( -7 sin(7y) ) / sqrt(.5cos(7y) ) OR (-7sin(7y))/(.5cos(7y))^.5
The .5 is in the wrong place. The derivative of \(\displaystyle (\space f(x)\space )^{\frac{1}{2}}\) is
\(\displaystyle \frac{1}{2}(\space f(x)\space )^{-\frac{1}{2}}=0.5\space (\space f(x)\space )^{-\frac{1}{2}}\)
NOT
\(\displaystyle (\space 0.5\space f(x)\space )^{-\frac{1}{2}}\)

4)(w^6)/(e^(w)sin(w))
((6w^5)(e^(w)sin(w))-(( w^6)(e^(w)cos(w))+(e^(w)sin(w))))/(e^(w)sin(w))^2
Quotient rule and Chain rule (sorry for the excessive amount of parenthesis, I am copying the answers I am typing into the computer and for some reason our software wasn't picking up terms correctly unless I had it like this.)
( (6w^5) ( e^(w) sin(w) ) - ( ( w^6 ) ( e^(w) cos(w) ) + ( e^(w) sin(w) ) ) ) / ( e^(w) sin(w) )^2
Remove the red close and open paren to get
\(\displaystyle \frac{6\space w^5\space e^w\space sin(w)\space -\space \{\space w^6\space [\space e^w\space cos(w)\space +\space e^w\space sin(w)\space ]\space \}
}{(\space e^w\space sin(w)\space )^2}\)

Thanks, both make sense, question though. Similar to the first one you did (number 3) when you do something like \(\displaystyle \ sin(3x)\) would it be \(\displaystyle \ 3(cos(3x)\) or \(\displaystyle \ (3cos(3x))\). Also if it is cos and you get -sin squared as the derivative would \(\displaystyle \ cos(3x)\) be \(\displaystyle \ 3(-sin(3x))\) or \(\displaystyle \ (-3sin(3x))\)? I am pretty sure it wouldn't be \(\displaystyle \ -3(sin(3x))\)

 

[Ishuda]

Thanks, both make sense, question though. Similar to the first one you did (number 3) when you do something like \(\displaystyle \ sin(3x)\) would it be \(\displaystyle \ 3(cos(3x)\) or \(\displaystyle \ (3cos(3x))\). Also if it is cos and you get -sin squared as the derivative would \(\displaystyle \ cos(3x)\) be \(\displaystyle \ 3(-sin(3x))\) or \(\displaystyle \ (-3sin(3x))\)? I am pretty sure it wouldn't be \(\displaystyle \ -3(sin(3x))\)

Looks like a typo with the red 'squared' above.

The derivative of cos(3x) can be written in the following equivalent forms
(cos(3x))' = 3(-sin(3x)) = (-3sin(3x)) = -3 (sin(3x)) = -3 sin(3x)

and similarly with the derivative of sin(3x)
(sin(3x))' = 3 (cos(3x)) = (3 cos(3x)) = 3 cos(3x)