Find the coefficient of the squared term in the simplified form for the second derivative., f"(x) for f(x)=(X^3+2x+3)(3x^3-6x^2-8x+1). use the hyphen symbol,-, for negative values.

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Find the coefficient of the squared term in the simplified form for the second derivative., f"(x) for f(x)=(X^3+2x+3)(3x^3-6x^2-8x+1). use the hyphen symbol,-, for negative values.

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Hey Maggy :)\[\large\rm f(x)=(x^3+2x+3)(3x^3-6x^2-8x+1)\]Hmm this one looks like a doozy. Remember your product rule?
\[\rm \color{royalblue}{f'(x)}=\color{royalblue}{(x^3+2x+3)'}(3x^3-6x^2-8x+1)+(x^3+2x+3)\color{royalblue}{(3x^3-6x^2-8x+1)'}\]This is how we would "set up" our first derivative, ya? :d
Alternatively, you could expand out all the brackets from the start. That might actually be easier :) Hmm I'm not sure.

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ya ya ya ya, expand first. that seems way better >.< Otherwise we have to combine a whole lot of term at the end.. my bad
we allowed to do that, ya? :d
my head hurts trying to follow you lol so we expand?
ya let's try that instead :) product rule will result in two groups of stuff. second derivative will require product rule twice more. then expanding three sets of things, and combining. expanding first seems way simpler
|dw:1438993951176:dw|
Here is the expansion of the first two terms from the first set of brackets multiplying everything in the other brackets. Does it kinda make sense what I'm doing? :d there is one more term to take care of.
yes I see you expanding to dind the derivatives. o.o math is not my forte lol
|dw:1438994231877:dw|we have another set of terms to get. so we'll have to distribute this 3.
|dw:1438994268041:dw|
We're going to take two derivatives. If we want to END UP WITH a second power. which power of x do you think we'll be starting with? :)
with 9x^3?
when we take the derivative of a third power such as that term, we'll get a second power. when we take the derivative again, it degrades down to first power, ya?
So it looks like we'll want to pay attention to the `fourth power` term. That is the one which becomes `second power` after differentiating it twice.
|dw:1438994442968:dw|
You `could` take the derivative of the entire thing. But it doesn't look the problem is requiring that, if I'm reading it correctly. They simply want the coefficient on the 2nd power term in your second derivative. So we currently have -2x^4. What do you get when you differentiate -2x^4 twice?
-2x^8??? ( when you say twice you mean add or multiply im getting very confused)
\(\large\rm (-2x^4)'=?\)
Remember your power rule? :)
sorta but not confidently
\(\large\rm (x^n)'=n x^{n-1}\) The exponent comes down in front to multiply, and the exponent decreases by 1.
Example: \(\large\rm (4x^7)'=4\cdot7x^{7-1}=28x^{6}\) See how the 7 comes down in front, and the power changes to a 6?
so it be 2x^3? from 2x^4?
Nooo :O Where did your 4 go? it didn't come down silly!
I cant see your last posted picture that wy I ask my answer was right lol all I see in your last picture posst is symbols
oh the math code didn't format correctly :( ugh
This was the example:|dw:1438995007902:dw|
The 7 comes down, and then the power decreases to a 6.
-8x^3? my reasoning: -2x^4= -2. 4x^3 = -8x3
Oo ok nice nice nice. That takes care of our first derivative. Now we have to repeat the process! :)|dw:1438995233431:dw|
-8x^3 -8x^3=-8.3x^2=-24x^2?
\[\checkmark\]
Yayyy good job \c:/ Notice that the exponent we started with was a 4. And the one we ended with is a 2. Differentiating twice made the power decrease by 2. That will happen to every other term as well. So our 5th power term will becomes a 3rd power, our 3rd power term will become a 1st power, and so on. So we only needed to differentiate the 4th power term twice, because that's the only one giving us 2nd power!
So yay good, you've found your coefficient! :) -24
ok so my answer to my question is -24x^2?
The question said `Find the coefficient`. The coefficient is `only` the number in front.
oh coolz hehe can you help me in one more question?
bahh i need a maf break :) open up a new thread so people can find your question faster! this one is a mess now hehe
lol ok ty

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