Product Rule

[Kingeivers]

Hey I am trying to solve this question but am stuck after the first differentiation.
F(x) = x(sqrt(3x-4)

It leaves me with,
Sqrt(3x-4) + 3x/2(sqrt(3x-4)
I know I have to simplify first before solving but am unsure on howto do so. Help would be muchly appreciated.

 

[Kingeivers]

I need help on question 18, I understand the question but cannot get past the 1st differentiation. Once I have differentiated it once, I am unable to then do it the second time. I assume their must be some simplyifing involved before going to f"(x). It has been bugging me so much and stressing me out. Instructions on where to go would be fantastic. Help a brother out, thankyou.

 

[MarkFL]

It's difficult to determine exactly what the radicand is...are we to interpret the given function as:

[MATH]f(x)=x\sqrt{3x-4}[/MATH] ?

 

[MarkFL]

I've merged the duplicate threads and given it a title that describes the question. It appears now that my interpretation is the function is correct.

You have correctly, from what I gather, determined:

[MATH]f'(x)=\sqrt{3x-4}+x\cdot\frac{3}{2\sqrt{3x-4}}[/MATH]
We may combine terms to obtain:

[MATH]f'(x)=\frac{9x-8}{2\sqrt{3x-4}}[/MATH]
To get the second derivative, apply the quotient rule...what do you get?

 

[Dr.Peterson]

Hey I am trying to solve this question but am stuck after the first differentiation.
F(x) = x(sqrt(3x-4)

It leaves me with,
Sqrt(3x-4) + 3x/2(sqrt(3x-4)
I know I have to simplify first before solving but am unsure on howto do so. Help would be muchly appreciated.

You are given f(x) = x sqrt(3x-4).

You have correctly differentiated, though the result should properly be written as f'(x) = sqrt(3x-4) + 3x/(2 sqrt(3x-4)).

There is really no simplification needed before the next differentiation; the second term is a quotient, to which you can apply the quotient rule. We'll want to see what you get when you try this, to see where you are getting tangled up.

I personally might write f'(x) as (3x-4)^(1/2) + (3/2)x (3x-4)^(-1/2). This way, the radicals are in the easier-to-work-with exponential form, and the second term can be done by the product rule, which I often find easier. But this is not necessary.

I see that MarkFL has suggested a simplification; although, again, that is not required, it will probably make the final simplification easier for you. Final simplification is not really necessary either; I've known some instructors to tell students to leave results unsimplified.

 

[MarkFL]

To follow up:

[MATH]f''(x)=\frac{2\sqrt{3x-4}(9)-(9x-8)\dfrac{3}{\sqrt{3x-4}}}{4(3x-4)}=\frac{3(9x-16)}{4(3x-4)^{\frac{3}{2}}}[/MATH]
Thus:

[MATH]f''(2)=\frac{3(9(2)-16)}{4(3(2)-4)^{\frac{3}{2}}}=\frac{3}{4\sqrt{2}}[/MATH]