anonymous
  • anonymous
y'' for y= (x^2+9)^4
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Take the first derivative using the chain rule and then take the derivative of that.
anonymous
  • anonymous
Second derivative will require both the chain rule and the product rule.
anonymous
  • anonymous
could you show me it worked step by step

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anonymous
  • anonymous
\[y'=4(x^2+9)^3(2x)\]
anonymous
  • anonymous
That's the first derivative. Do you see where things might have come from?
anonymous
  • anonymous
yes i had it all right except the 2x and thats by taking the derivative of the things inside the brackets
anonymous
  • anonymous
Yes :) that is the chain rule.
anonymous
  • anonymous
So now we move on to the second derivative which involves the product rule.
anonymous
  • anonymous
Do you know how to use the product rule?
anonymous
  • anonymous
okay i got.... 4(x^2+9)^3 (2) + (2x) 12(x^2+9)^2
anonymous
  • anonymous
Very close. First half is correct. Second half you forgot something.
anonymous
  • anonymous
what did i forget
anonymous
  • anonymous
oh do the product rule to finish?
anonymous
  • anonymous
the books has 8(x^2+9)^3 + 48x^2(x^2+9)^2 = 8(x^2+9)^2(7x^2+9)
anonymous
  • anonymous
Well the book is correct lol. All you forgot in your was the chain rule on the second half. Should have been (2x)12(x^2+9)(2x) that 2x on the end is what you forgot.
anonymous
  • anonymous
why is the 2x at the end as well?
anonymous
  • anonymous
Ok so when you did the second half of the product rule you can say to yourself (2x) times the derivative of the function. Well the derivative of that function 4(x^2+9)^3 is 12(x^2+9)^(2) times the chain rule (2x).
anonymous
  • anonymous
okay so now where do they get their answer
anonymous
  • anonymous
got the first part, but didnt get it after the simplifying
anonymous
  • anonymous
They probably multiplied stuff out and then it eventually got them to that answer. Truthfully, I have never seen a teacher who would not accept the first answer you have. The other answer is just tedious work.
anonymous
  • anonymous
okay thanks!
anonymous
  • anonymous
Actually I just looked at their second answer and you could factor...
anonymous
  • anonymous
okay could you help me with a ln y''?
anonymous
  • anonymous
Sure.
anonymous
  • anonymous
okay. I am horrible at ln problems but here is the original y'' for y= ln x/x^2
anonymous
  • anonymous
Ok this is a quotient rule problem or can be a product rule if you rearrange it. Take your pick.
anonymous
  • anonymous
SO IT COULD BE (LN X)(-X^2)
anonymous
  • anonymous
Well could be lnx(x^-2). Normally a lot of people prefer to use the product rule when possible, but it's a little easier.
anonymous
  • anonymous
OH OKAY
anonymous
  • anonymous
what is ln x '
anonymous
  • anonymous
The derivative of lnx is 1/x
anonymous
  • anonymous
oh yeah sorry
anonymous
  • anonymous
No worries :)
anonymous
  • anonymous
okay so y' is lnx(-2x^-3) + (x^-2)(1/x)
anonymous
  • anonymous
Yes. :)
anonymous
  • anonymous
okay now im going to need some help..lol
anonymous
  • anonymous
With simplifying?
anonymous
  • anonymous
no just straight so i can see what to do
anonymous
  • anonymous
I'm confused. So what do you need help on? Sorry lol
anonymous
  • anonymous
second derivative
anonymous
  • anonymous
Ah. Ok. Well you can use the product rule again for the second derivative, but twice. So use the product rule for the lnx(-2x^-3) and then add that with the result of the product rule of (x^-2)(x^-1). I rewrote 1/x so it's easier to see.
anonymous
  • anonymous
okay cool
anonymous
  • anonymous
can you show me it worked out?
anonymous
  • anonymous
Here is the working

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