y'' for y= (x^2+9)^4

- anonymous

y'' for y= (x^2+9)^4

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- anonymous

Take the first derivative using the chain rule and then take the derivative of that.

- anonymous

Second derivative will require both the chain rule and the product rule.

- anonymous

could you show me it worked step by step

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## More answers

- anonymous

\[y'=4(x^2+9)^3(2x)\]

- anonymous

That's the first derivative. Do you see where things might have come from?

- anonymous

yes i had it all right except the 2x and thats by taking the derivative of the things inside the brackets

- anonymous

Yes :) that is the chain rule.

- anonymous

So now we move on to the second derivative which involves the product rule.

- anonymous

Do you know how to use the product rule?

- anonymous

okay i got....
4(x^2+9)^3 (2) + (2x) 12(x^2+9)^2

- anonymous

Very close. First half is correct. Second half you forgot something.

- anonymous

what did i forget

- anonymous

oh do the product rule to finish?

- anonymous

the books has 8(x^2+9)^3 + 48x^2(x^2+9)^2 = 8(x^2+9)^2(7x^2+9)

- 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

why is the 2x at the end as well?

- 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

okay so now where do they get their answer

- anonymous

got the first part, but didnt get it after the simplifying

- 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

okay thanks!

- anonymous

Actually I just looked at their second answer and you could factor...

- anonymous

okay could you help me with a ln y''?

- anonymous

Sure.

- anonymous

okay. I am horrible at ln problems but here is the original
y'' for y= ln x/x^2

- anonymous

Ok this is a quotient rule problem or can be a product rule if you rearrange it. Take your pick.

- anonymous

SO IT COULD BE (LN X)(-X^2)

- 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

OH OKAY

- anonymous

what is ln x '

- anonymous

The derivative of lnx is 1/x

- anonymous

oh yeah sorry

- anonymous

No worries :)

- anonymous

okay so y' is lnx(-2x^-3) + (x^-2)(1/x)

- anonymous

Yes. :)

- anonymous

okay now im going to need some help..lol

- anonymous

With simplifying?

- anonymous

no just straight so i can see what to do

- anonymous

I'm confused. So what do you need help on? Sorry lol

- anonymous

second derivative

- 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

okay cool

- anonymous

can you show me it worked out?

- anonymous

Here is the working

##### 1 Attachment

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