Finding the derivative and expressing in factored form

[meeklobraca]

Homework Statement

Find the derivative of y = (x-2)exp3 x sqrt 2x-1

Homework Equations

The Attempt at a Solution

I got a final answer of (x-2)^2(2x-1)^-1/2(13x-14)

What do you guys think? Is this correct?

 

[Defennder]

I can't read the original question. Is it supposed to be: y=(x-2)e^{3\sqrt{2x-1}}?

 

[meeklobraca]

Yeah that's supposed to be the question. I don't know how to type it up any better than that sorry.

 

[Defennder]

Well I realized I can't read your answer as well. But I don't think it's correct since the exponential term doesn't appear anywhere.

 

[meeklobraca]

Sorry, I used the ^ for the exponent

It should read (x-2)exp2(2x-1)exp-1/2(13x-14)

 

[meeklobraca]

Again I am sorry, i made a mistake. The question should read

y = ((x-2)exp3) (sqrt 2x-1)

 

[FedEx]

Again I am sorry, i made a mistake. The question should read

y = ((x-2)exp3) (sqrt 2x-1)

Take log on both sides and then try differentiating it

 

[HallsofIvy]

Still ambiguous.
y= (x-2)^3 \sqrt{2x-1}
or y= (x-2)^3(\sqrt{2x}-1)?

If the first, write it as (x-2)^3 (2x-1)^{1/2} and use the chain rule and product rule.

If the second, write it as (x-2)^3(\sqrt{2}x^{1/2}- 1) and again use the chain rule and product rule.

 

[meeklobraca]

What is the definition of the chain rule with the product rule? They give an example of that in the manual but its not very clear.

 

[HallsofIvy]

The product rule says
d(uv)/dx = (du/dx)v+ u(dv/dx)

The chain rule is d(u(v))/dx= du(v)/dv (dv/dx)

 

[meeklobraca]

How come I have to use both of those rules? Wouldnt 1 suffice?

 

[kidmode01]

Applying the product rule first you realize you'll be using the chain rule automatically when calculating the derivatives: Take the derivative of the first term,expression,etc and multiply it by the second term. Thats the first part of your product rule. Then ADD: the derivative of the second term multiplied by the first term. (Or whichever order you prefer)

(f*g)' = f'*g + g'*f

The ' , just means the derivative. f' means the derivative of f, and g' means the derivative of g.

Now for the chain rule. When you take the derivative of the first term you notice you need to take the derivative of (x-2)^3, bring the 3 down infront, minus 1 from the exponent, and multiply by the derivative of the inside of (x-2), which is 1:

3(x-2)^2 * 1

A similar operation will be done for the next derivative, if you have problems seeing the "power rule/chain rule" with the square root, just change the squareroot to the power of 1/2. The chain rule is used when you have functions nested inside other functions. So it need not be a power. Say you have cos(5x), the 5x is nested inside the cos(). So taking the derivative you would get -sin(5x)*5. Generally it's just a good idea to remember the power rule and the chain rule applied to trigonomic functions. If you need more examples or maybe another way of describing it, just google power rule or chain rule examples or search the forums. If you really want to know where it comes from, try picking up an introductory analysis textbook or just search for a proof.

Quick example, try doing this one first. Its a little easier: The derivative of:
(5x-1)^2 (4+2x)^3<br /> =(5x-1)^2*3(4+2x)^2 2 + (4+2x)^3 *2(5x-1) 5

 

[meeklobraca]

Do I always use the product rule, followed by the chain rule when finding the deriviative of 2 factors?

 

[NoMoreExams]

Nope, if you had to differentiate cos(x)*e^x you can just use the product rule (you are still using the chain rule when you are taking the derivative of the arguments of your function but since they're just x and you know the derivative of x is 1 and multiplying by 1 is ... well you get the idea)

 

[meeklobraca]

OKay, so let's work this through with the question I have from above.

I got (2x-1)^-1/2(x-2)^2(7x-5) for a final answer?

 

[Jamil (2nd)]

That's wrong unfortunately. Use the chain rule on (x-1)³ and (2x-1)^1/2 separately. The combine them and use the product rule.

e.g by using chain rule: (d/dx)(x-1)³ = 3(x-1)²

 

[meeklobraca]

The first factor should be (x-2)^3

Okay, I used the chain rule on the 2 factors separately. I got 3(x-2)^2 and (2x-1)^-1/2

Now I don't knoow how to use the product rule with factors that have exponents in them.

Using the product rule I get an equation of 3(x-2)^2 -1/2(2x-1)^-3/2 + (2x-1)^-1/2 6(x-2)

 

[meeklobraca]

Im still having trouble with this question here.

Using the product rule and the chain rule together I have

3(x-2)^2(2x-1)^1/2 + (2x-1)^-3/2 (x-2)^3

so from here I need to factor out some of this,

(x-2)^2(2x-1)^-3/2 [x-2/3(2x-1)^1/3


where do i go from here?

 

[NoMoreExams]

I'll work this out for you trying to detail each step, usually I'm in favor of suggesting hints rather than giving solutions but it has been long enough where you might benefit more from reading each step in detail.

All right assuming this is the problem (x-2)^3 \sqrt{2x-1}.

Basically we have a product of 2 functions

f(x) = (x-2)^3 and g(x) = \sqrt{2x-1}.

We know the product rule for derivatives tells us:

(f(x) g(x))^{&#039;} = f^{&#039;}(x)g(x) + f(x)g^{&#039;}(x)

(where ' denotes derivative)

So in our case we get the following:

f(x) = (x-2)^3 \Rightarrow f&#039;(x) = 3(x-2)^2 and g(x) = \sqrt{2x-1} \Rightarrow g&#039;(x) = \frac{1}{\sqrt{2x-1}}

Putting this together into our formula we get:

3(x-2)^{2}\sqrt{2x-1} + \frac{(x-2)^{3}}{\sqrt{2x-1}} = \frac{(x-2)^{2}}{\sqrt{2x-1}}\left(3(2x-1) + (x-2)\right) = \frac{(x-2)^{2}}{\sqrt{2x-1}}(7x-5)

Personally I have found that most Calculus students suffer not because they do not understand the Calculus portion of the material but because their Algebra is rather spotty. It might help in your studies if you went back and reviewed some of the old material.

 

[meeklobraca]

Thats kind of frustrating cause I got that as the final answer in one of my previous posts, with the exception of not bringing the 2x-1^-1/2 down, and the one guy said that was wrong and I should try it another way.

 

[Saladsamurai]

Thats kind of frustrating cause I got that as the final answer in one of my previous posts, with the exception of not bringing the 2x-1^-1/2 down, and the one guy said that was wrong and I should try it another way.

I see that in post #15; however, if you are not going to use the LaTex templates, you need to be careful to use parentheses properly because your notation in that post is ambiguous (though if Jamil indeed knew the correct answer to that, they should have been able to decipher your notation).

 

[meeklobraca]

Its all good, the positives of the help provided here far outweigh anything else. Thank you NME for the help.