Find my flaws in derivative solving, be my light in the dark!

[animalplanet]

I am doing a chapter review in calculus and I'm stuck on a few problems. I'm going to post the work I have done so far for them, and it'd be great if someone could help me out with any of them. Doesn't have to be all, just some extra help wherever you feel like helping would be nice. I would really appreciate guidance on what to do.


1) Use analytic methods to find the global extreme values of the function and state where they occur.
y= x sqroot (2-x)
y= x (2-x)^^(1/2)
Find first derivative and set equal to zero
y'= 1 + (2-x^1/2) / 2x (2-x^3/2)



3) For 3, 11, 13, and 15, find the intervals where the function is increasing, decreasing, concaving up, and concaving down. Find any local extreme values and inflection points.
y= (x²)(e^ ^1/x2)
Find first derivative using product rule.
u' = 2x
v' = This is where I get stuck, finding the derivative of e


11) y= ln |x|
Simplified to 1 / |x|
Find first derivative
y' = 1/x
u' = 1
v' = 1
I start to get confused here. I think I have a problem finding derivatives of some equations


13) This is a pair of two equations in a function
y = e ^-x
y = 4x - x³

y' = -e ^-x
y' = 4 - 3x²

Now I get confused on what x equals.


15)
y = x^4/5 (2-x)
u' = 4/5 x^-1/5
v' = 1
y' = (x^4/5)(1) + (4/5x^-1/5)(2-x)
y' = 9/5 x^4/5 + 8/5 x^-1/5

Again I get lost now trying to find the extreme values, inflection points etc



19) Find all possible functions with the given derivative. (I'm not quite sure what it's asking me to do)
f'(x) = x^-5 + e^-x



21) Find all possible functions with the given derivative.
f'(x) = 2/x + x² + 1


40) Let y = x²e^-x
Find dy and evaluate dy for x = 1 and dx = 0.01
When finding dy I get confused. I feel like my main problem is finding the derivatives sometimes

 

[MarkFL]

1.) You have differentiated incorrectly. Given:

$\displaystyle y=x(2-x)^{\frac{1}{2}}$

We find using the product rule:

$\displaystyle \dfrac{dy}{x}=x\cdot\dfrac{d}{dx}\left((2-x)^{\frac{1}{2}} \right)+\dfrac{d}{dx}(x)\cdot(2-x)^{\frac{1}{2}}=$

$\displaystyle x\cdot\dfrac{-1}{2\sqrt{2-x}}+\sqrt{2-x}=\dfrac{-x+2(2-x)}{2\sqrt{2-x}}=\dfrac{4-3x}{2\sqrt{2-x}}$

The given function is defined on $\displaystyle (-\infty,2]$. We see the function is increasing on $\displaystyle \left(-\infty,\dfrac{4}{3} \right)$ and decreasing on $\displaystyle \left(\dfrac{4}{3},2 \right)$. So what kind of extremum do we have at our critical value?

edit: I like to take problems one at a time, so once we have solved the first one to your satisfaction, we can move on to the second, if no one else has posted on it. If someone does, then I will let them handle that one with you without interfering.

 

[HallsofIvy]

3)

For 3, 11, 13, and 15, find the intervals where the function is increasing, decreasing, concaving up, and concaving down. Find any local extreme values and inflection points.

y= (x²)(e^ ^1/x2)
Find first derivative using product rule.
u' = 2x
v' = This is where I get stuck, finding the derivative of e

You should know that the derivative of $\displaystyle e^x$ is just $\displaystyle e^x$ again. And then if u(x) is any differentiable function of x, by the chain rule, the derivative of $\displaystyle e^{u(x)}$ is $\displaystyle e^{u(x)}\frac{du}{dx}$.

11)

y= ln |x|
Simplified to 1 / |x|

I don't know what you mean by "simplified". I assume you mean that the derivative of y= ln|x| is 1/|x|. That is almost correct. If x> 0 then |x|= x and the derivative is 1/x which, because x> 0, is the same as 1/|x|. If x< 0, then |x|= -x so ln|x|= ln(-x) and. by the chain rule, the derivative is (1/(-x))(-1)= 1/x. Here, x is NOT positive so 1/x is NOT 1/|x|. 1/|x| is not the derivative of ln|x|, 1/x is.

Find first derivative
y' = 1/x

Yes, as I said above.

u' = 1
v' = 1
I start to get confused here. I think I have a problem finding derivatives of some equations

I have no clue what you are doing! Where did "u": and "v", much less u' and v', come from? You had already solved the problem- the derivative of ln|x| is 1/x as you said in your first line.

13) This is a pair of two equations in a function
y = e ^-x
y = 4x - x³

y' = -e ^-x
y' = 4 - 3x²

Now I get confused on what x equals.

What is the question? You are given two equations. I'm not clear what you mean by "in a function". You differentiated as if you had two different functions. Were these two components of a "vector valued" function? And why should x equal any specific number. It's impossible to tell what you are asking because you have not stated exactly what the original question was.

15)
y = x^4/5 (2-x)
u' = 4/5 x^-1/5
v' = 1
y' = (x^4/5)(1) + (4/5x^-1/5)(2-x)
y' = 9/5 x^4/5 + 8/5 x^-1/5

Again I get lost now trying to find the extreme values, inflection points etc

Please, please, please, tell us what you are doing! Writing "u' = 4/5 x^-1/5" without saying what u is, is confusing! I assume that you are trying to do this using the product rule with u= x^4/5 and v= 2- x. Yes, the derivative of u is u'= (4/5)x^-1/5 but the derivative of v is v'= -1, not 1. You could also do this proble, perhaps more easily by using the "distributive law" at the first: y= x^4/5(2- x)= 2x^4/5- x^9/5 so y' = (4/5) x^-1/5- (9/5)x^4/5.

For a differentiable function, extreme values occur where the derivative changes sign which can only happen where the derivative is 0 so you need to start by solving y' = (4/5) x^-1/5- (9/5)x^4/5= 0 for x. But be careful. While being equal to 0 is necessary to changing sign, it is not sufficient. It is possible for the derivative to go to 0, then continue with the sae sign. In that case, the point is an "inflection point".

To solve (4/5) x^-1/5- (9/5)x^4/5= 0, multiply on both sides by x^1/5. That makes the equation (4/5)- (9/5)x= 0. Writing it as (-1/5)(9x- 4)= 0 makes it easy not only to find where the derivative is 0 but whether or not it changes sign there.

 

[lookagain]

I am doing a chapter review in calculus and I'm stuck on a > > > few < < < problems.

animalplanet,

eight problems is not a "few" problems. Please start a separate thread for each problem.

The first paragraph with a lead sentence in bold addresses this in the link below:


http://www.freemathhelp.com/forum/threads/41538-Read-Before-Posting!!

 

[animalplanet]

animalplanet,

eight problems is not a "few" problems. Please start a separate thread for each problem.

The first paragraph with a lead sentence in bold addresses this in the link below:


http://www.freemathhelp.com/forum/threads/41538-Read-Before-Posting!!

I'm wholeheartedly sorry. You're completely right. I didn't realize when I was writing out the problems I was having trouble with that there were so many. The whole review I was working on was over 50 problems, so as I was going through the ones I did not entirely understand I didn't realize the number of problems I was listing on the forum. I sincerely apologize and I will never neglect this rule again.

 

[animalplanet]

1.) You have differentiated incorrectly. Given:

$\displaystyle y=x(2-x)^{\frac{1}{2}}$

We find using the product rule:

$\displaystyle \dfrac{dy}{x}=x\cdot\dfrac{d}{dx}\left((2-x)^{\frac{1}{2}} \right)+\dfrac{d}{dx}(x)\cdot(2-x)^{\frac{1}{2}}=$

$\displaystyle x\cdot\dfrac{-1}{2\sqrt{2-x}}+\sqrt{2-x}=\dfrac{-x+2(2-x)}{2\sqrt{2-x}}=\dfrac{4-3x}{2\sqrt{2-x}}$

Thank you so much for clarifying that! I think I easily get confused with differentiating. Now that you've pointed out my issue there solving the rest makes a lot more sense.

 

[phoenix9124]

19) Find all possible functions with the given derivative. (I'm not quite sure what it's asking me to do)
f'(x) = x^-5 + e^-x



21) Find all possible functions with the given derivative.
f'(x) = 2/x + x² + 1

Let's say you have the function:
f(x) = 3x^2 + 9
If you differentiate this, you get:
f'(x) = 6x
Now if you're only been given f'(x) = 6x
you don't know that there should also be a +9 in there.
Since constants go away once you differentiate, you need to add a constant when you go the other way around.
So in this case it would be:
f(x) = 3x^2 + c
With c being any number unless other conditions are given that makes it possible to calculate c ( for example, point A(2,3) is on the graph )