Finding the Second Derivative of a Square Root Function

[illegalvirus]

Homework Statement

Find the exact value of f''(2) if f(x)=\sqrt{3x-4}

Homework Equations

See above

The Attempt at a Solution

I've tried to use the product rule to differentiate.

f= x(3x -4)^{\frac{1}{2}}

f'= (3x -4)^{\frac{1}{2}} + \frac{3}{2}^{\frac{-1}{2}}

f''= \frac{3}{2}x(3x -4)^{\frac{-1}{2}} . \frac{3}{2}x(3x -4)^{\frac{-1}{2}} + (3x -4)^{\frac{1}{2}} . \frac{-9}{4}x(3x-4)^{\frac{-3}{2}}

= \sqrt{3x -4} . \frac{-9x}{4\sqrt{(3x-4)^{3}}}

Now I am kind of stuck here. I've tried it a few times and I have no idea what I am doing wrong, some advice please?

 

[Anti-Meson]

Use the chain rule instead

 

[illegalvirus]

Somehow I got the same answer

f'(x) = \frac{3}{2}x(3x -4)^{\frac{-1}{2}}
f''(x) = \frac{-9}{4}(3x -4)^{\frac{-3}{2}}

 

[Anti-Meson]

Somehow I got the same answer

f'(x) = \frac{3}{2}x(3x -4)^{\frac{-1}{2}}
f''(x) = \frac{-9}{4}(3x -4)^{\frac{-3}{2}}

Your first derivative is incorrect. However, you have somehow ended up with the correct second derivative :S.

f''(x) = \frac{-9}{4}(3x -4)^{\frac{-3}{2}}[BTW: f'(x) = \frac{3}{2}(3x -4)^{\frac{-1}{2}} ]

 

[Mark44]

Homework Statement

Find the exact value of f''(2) if f(x)=\sqrt{3x-4}

Homework Equations

See above

The Attempt at a Solution

I've tried to use the product rule to differentiate.

f= x(3x -4)^{\frac{1}{2}}

Why are you writing your function like this? There is no factor of x outside the radical, so the product rule is not called for here. This is your function.
f(x)=\sqrt{3x-4}

f'= (3x -4)^{\frac{1}{2}} + \frac{3}{2}^{\frac{-1}{2}}

f''= \frac{3}{2}x(3x -4)^{\frac{-1}{2}} . \frac{3}{2}x(3x -4)^{\frac{-1}{2}} + (3x -4)^{\frac{1}{2}} . \frac{-9}{4}x(3x-4)^{\frac{-3}{2}}

= \sqrt{3x -4} . \frac{-9x}{4\sqrt{(3x-4)^{3}}}

Now I am kind of stuck here. I've tried it a few times and I have no idea what I am doing wrong, some advice please?

 

[illegalvirus]

Sorry, I think the first derivative is this,

f'(x) = \frac{1}{2}x(3x -4)^{\frac{-1}{2}}

 

[Anti-Meson]

Please read your textbook on chain rule. It is wrong to put an x in the front or a half for that matter.

 

[illegalvirus]

Why are you writing your function like this? There is no factor of x outside the radical, so the product rule is not called for here. This is your function.
f(x)=\sqrt{3x-4}

Ahh, whoops. The original equation to derive is actually f(x)=x\sqrt{3x-4}!

 

[Anti-Meson]

Ok, in that case you were right to use the product rule.

f'(x) = \frac{3x}{2}(3x-4)^{-0.5}+(3x-4)^{0.5}

I leave the rest to you.

 

[illegalvirus]

Ok, in that case you were right to use the product rule.

f'(x) = \frac{3x}{2}(3x-4)^{-0.5}+(3x-4)^{0.5}

I leave the rest to you.

Wait, how did you get that?

 

[Anti-Meson]

Product rule.
\frac{d[u(x)v(x)]}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}

 

[illegalvirus]

Okay thank you, I've just realized that I've forgotten to multiply by the derivative of the brackets.

 

[Anti-Meson]

Just in case you get stuck the answer is:

3(3x-4)^{-0.5}-\frac{9}{4}x(3x-4)^{-1.5}

Though you should work this out by yourself. To verify.

 

[illegalvirus]

Just in case you get stuck the answer is:

3(3x-4)^{-0.5}-\frac{9}{4}x(3x-4)^{-1.5}

Though you should work this out by yourself. To verify.

I got this, \frac{3}{2}(3x-4)^{-0.5}-\frac{9}{4}x(3x-4)^{-1.5}

But when finding the exact value of f''(2) I had a bit of a problem as well.

So far I have, f''(2)= \frac{-9}{4\sqrt{2}} + \frac{3}{2\sqrt{2}}.

=\frac{-6\sqrt{2}}{4\sqrt{2}}

But the answer is \frac{3\sqrt{2}}{8}

 

[Anti-Meson]

That answer arises from my answer (i.e. the correct answer) go through your derivation again to check the mistakes.