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.