Help on finding 2nd derivative

[Jacob]

Can someone please check my workings and see where am I making mistake when I'm finding f"(x).

 

[JeffM]

First off, your second line is technically wrong. You mean

[MATH]f'(x) = - 3 \left ( - \dfrac{1}{2} * (3x^2 + 3)^{-3/2} * 6x \right) \implies[/MATH]
[MATH]f'(x) = 9x(3x^2 + 3)^{-3/2}.[/MATH]
But to find the derivative, you must use the product rule. You used only half the product rule.

Second, thank you for showing your work in detail and so clearly.

 

[Jacob]

First off, your second line is technically wrong. You mean

[MATH]f'(x) = - 3 \left ( - \dfrac{1}{2} * (3x^2 + 3)^{-3/2} * 6x \right) \implies[/MATH]
[MATH]f'(x) = 9x(3x^2 + 3)^{-3/2}.[/MATH]
But to find the derivative, you must use the product rule. You used only half the product rule.

Second, thank you for showing your work in detail and so clearly.

Yes, that's how I get [MATH]f'(x) = 9x(3x^2 + 3)^{-3/2}.[/MATH]
May I know from where I should use the product rule? Thank you.

Edit: I think I see where I did it wrong

 

[Jacob]

I see how my second line is wrong and I did the same mistake when I was trying to find the second derivative. Thats how I ended up getting -81x² as my numerator.

As you can see from my workings below, I still don't understand how to get to the final answer as -54x²+27 when I have positive 54x² and - 27x with the denominator with 2

 

[JeffM]

My answer was unclear. Let’s go back.

You correctly found the first derivative. Your technical mistake was in the second line of its derivation

When you wrote

[MATH]f(x) = -3(3x^2 + 3)^{-1/2}\\ \ \ \ \ \ \ \ = -3 \left \{- \dfrac{1}{2} * (3x^2 + 3)^{-3/2} * 6x \right \}[/MATH]That means

[MATH]f(x) = -3(3x^2 + 3)^{-1/2}\\ f(x) = -3 \left \{- \dfrac{1}{2} * (3x^2 + 3)^{-3/2} * 6x \right \}[/MATH]And that is incorrect. But what you meant was

[MATH]f(x) = -3(3x^2 + 3)^{-1/2}\\ f’(x)= -3 \left \{- \dfrac{1}{2} * (3x^2 + 3)^{-3/2} * 6x \right \}[/MATH]Which is correct.

When you leave the left hand side of an equation blank, that means it is UNCHANGED from the previous line. It implies ditto marks. You meant to change f(x) to f’(x) so you cannot imply ditto marks.

It was a purely technical mistake in NOTATION. But you correctly found the first derivative. Understand?

And then you simplified the first derivative. Either

[MATH]9x(3x^2 + 3)^{-3/2} \text { or } \dfrac{9x}{(3x^2 + 3)^{3/2}}[/MATH]
is a perfectly acceptable simplification.

Now if you use the first form, taking the second derivative requires
the product rule with respect to [MATH]9x \text { and } (3x^2 + 3)^{-3/2}.[/MATH]
If you use the second form, taking the second derivative obviously requires the quotient rule.

It looks as though you decided to use the first form for finding the second derivative, but neglected to use the product rule.

 

[Jacob]

The part where I stuck right now is how to solve it when I get it here to get the final answer where -54x²+27/(3x²+3)^5/2. I'm not familiar with solving terms that have power by using the Product rule or Quotient Rule.

May I know how I can solve this after I got to these steps?

 

[JeffM]

Have you been taught substitution of variables? If not, you soon will. Here we go.

[MATH]f'(x) = g(x) = 9x(3x^2 + 3)^{-3/2} \implies f''(x) = g'(x).[/MATH]
WIth me so far?

[MATH]Let u = 9x \text { and } v = (3x^2 + 3)^{-3/2} \implies[/MATH]
[MATH]g(x) = u * v.[/MATH]
Using these substituted variables, it is easy to use the product rule.

[MATH]g'(x) = u * v' + u' * v.[/MATH]
Any questions?

[MATH]u = 9x \implies u' = 9 \implies g'(x) = 9x * v' + 9 * v.[/MATH]
Easy peasy.

[MATH]v = (3x^2 + 3)^{-3/2} \implies v' = - \dfrac{3}{2} * (3x^2 + 3)^{-5/2} * 6x.[/MATH]
Very similar to what you did on the first derivative.

[MATH]\therefore g'(x) = 9x * \left ( - \dfrac{3}{2} * (3x^2 + 3)^{-5/2} * 6x \right ) + 9 * (3x^2 + 3)^{-3/2}) =[/MATH]
[MATH]9(3x^2 + 3)^{-5/2}\{-9x^2 + (3x^2 + 3)^{\{(-3/2)-(-5/2)\}}) =[/MATH]
[MATH]9(3x^2 + 3)^{-5/2}\{-9x^2 + (3x^2 + 3)^{2/2}\} =[/MATH]
[MATH]9(3x^2 + 3)^{-5/2}\{-9x^2 + (3x^2 + 3)^1\} =[/MATH]
[MATH]9(3x^2 + 3)^{-5/2}(-9x^2 + 3x^2 + 3) =[/MATH]
[MATH]\dfrac{9(-6x^2 + 3)}{3x^2 + 3)^{5/2}} \implies[/MATH]
[MATH]g'(x) = - \dfrac{54x^2 - 27}{3x^2 + 3)^{5/2}} = f''(x).[/MATH]
Simple concepts; lots of ugly algebra.

 

[Jacob]

Thank you.