Question

How to find the derivative of y=(x^2-3)^2/sqr 4x+5

Answer 1

you know the quotient law?

Answer 2

ok, what steps have you taken to try and solve this so far?

Answer 3

chain and quotient rule. But i'm having problems with the factoring thats all, not the rules

Answer 4

ok, well let me run through it really quickly

Answer 5

ok

Answer 6

the book fully factors the numerator of the answer?

Answer 7

yes

Answer 8

\[y=\frac{(x^2-3)^2}{\sqrt{4x+5}}\\y'=\frac{2(2x)(x^2-3)\sqrt{4x+5}-(x^2-3)^2(\frac{1}{2})(4x+5)^{\frac{-1}{2}}(4)}{4x+5}\\=4(4x+5)^{\frac{-1}{2}}\frac{2x(x^2-3)(4x+5)-(x^2-3)}{2(4x+5)}\\=2\frac{8x^4+10x^3-25x^2-30x-3}{(4x+5)^{\frac{3}{2}}}\]

Answer 9

wow that took 7 minutes. anyways that's as far as id go... really hope I didn't make a mistake lol

Answer 10

when you multiply by the bottom will it be -1/2 or 1/2 since its in the denominator

Answer 11

+1/2

Answer 12

is that the answer in the book?

Answer 13

the book has 2(x^2-3)(7x^2+10x+3) over the same denominator you have

Answer 14

hm I made an algebraic error somewhere, but look at the second last line, where I have (x^2-3) as a factor?

Answer 15

i don't see a mistake. but in the second last line, you factor out the (x^2-3) and you are left with \[y'=\frac{2(x^2-3)(8x^2+10x-1)}{(4x+5)^{\frac{3}{2}}}\]

Answer 16

ummm can you explain what you did when you factored out the sqr of 4x+5, because i'm still confused with that. Shouldn't the 4x+5 in the equation now have a -1 as an exponent ?

Answer 17

so when you have \[5a+2a^{-1}\\=a^{-1}(5a^2+2)\] you take out the one with the lowest exponent value. so in this example, its a^(-1) and just think about the exponent laws, and what the exponents inside the brackets shoulw be so that when you multiply the a^(-1) back in hen you get the original EQ again

Answer 18

but in the first part of the equation you take out a -1/2 and are left with a positive 1 as an exponent, but if you multiply it back you will get a negative 1/2, which isn't what the original was

Answer 19

ah but then you multiply the (4x+5)^-0.5 by the -(x^2-3) and are back to the first line

Answer 20

genius, do you see a mistake in my math?

Answer 21

\[(4x+5)^{\frac{-1}{2}}[(4x+5)-1]\\=(4x+5)^{\frac{1}{2}}-(4x+5)^{\frac{-1}{2}}\]

Answer 22

0213 you get what i did?

Answer 23

yes

Answer 24

\[y=\frac{ (x^2-3)^2 }{ \sqrt{4x+5} }\]\[y'=\frac{ 2(x^2-3)(2x)(\sqrt{4x+5})-\frac{ 1 }{ 2 }(4x+5)^{-\frac{ 1 }{ 2 }}(4)(x^2-3)^2 }{ 4x+5 }\]\[=\frac{ (4x^3-12x)(\sqrt{4x+5})-\frac{ (4)(x^2-3)^2 }{ 2\sqrt{4x+5} } }{ 4x+5 }=\frac{ \frac{ 2\sqrt{4x+5}(4x^3-12x)(\sqrt{4x+5}) }{ 2\sqrt{4x+5} }-\frac{ (4)(x^2-3)^2 }{ 2\sqrt{4x+5} } }{ 4x+5 }\]\[=\frac{ \frac{ 2(4x+5)(4x^3-12x) }{ 2\sqrt{4x+5} }-\frac{ (4)(x^2-3)^2 }{ 2\sqrt{4x+5} } }{ 4x+5 }=\frac{ \frac{ [2(4x+5)(4x^3-12x)]-[4(x^2-3)^2] }{ 2\sqrt{4x+5} } }{ 4x+5 }\]\[=\frac{ \frac{ [32x^4-96x^2+40x^3-120x]-[4(x^4-6x^2+9)] }{ 2\sqrt{4x+5} } }{ 4x+5 }=\frac{ \frac{ [32x^4-96x^2+40x^3-120x]-[4x^4-24x^2+36] }{ 2\sqrt{4x+5} } }{ 4x+5 }\]\[=\frac{ \frac{ 32x^4-96x^2+40x^3-120x-4x^4+24x^2-36 }{ 2\sqrt{4x+5} } }{ 4x+5 }=\frac{ \frac{ 28x^4-72x^2+40x^3-120x-36 }{ 2\sqrt{4x+5} } }{ 4x+5 }\]\[=\frac{ \frac{4( 7x^4-18x^2+10x^3-30x-9) }{ 2\sqrt{4x+5} } }{ 4x+5 }=\frac{ \frac{2( 7x^4-18x^2+10x^3-30x-9) }{ \sqrt{4x+5} } }{ 4x+5 }\]\[= \frac{2( 7x^4-18x^2+10x^3-30x-9) }{ \sqrt{4x+5} } \div \frac{ 4x+5 }{ 1 }\]\[= \frac{2( 7x^4+10x^3-18x^2-30x-9) }{ \sqrt{4x+5} } \times \frac{ 1 }{ 4x+5 }=\frac{2( 7x^4+10x^3-18x^2-30x-9) }{ \sqrt{4x+5}(4x+5) } \]\[=\frac{2( 7x^4+10x^3-18x^2-30x-9) }{ (4x+5)^{\frac{ 3 }{ 2 }} } \rightarrow y'=\frac{2( 7x^4+10x^3-18x^2-30x-9) }{ \sqrt[]{(4x+5)^3} }\]
@0213

Took me forever. Hope this helps.

Answer 25

@matineesuxxx

Answer 26

omg thank you.

Answer 27

lol ya I skipped a few steps. thanks :)

Answer 28

It looks so damn beautiful. I'm gonna keep looking at that the whole day lol.

Answer 29

haha and that does factor out to what he needed

Answer 30

What do you mean factor out? The final derivative is in the last row after y'=

Answer 31

ya but his books factors the numerator

Answer 32

No, if you factor the top, nothing cancels out. So it's best to leave it unfactored like that. You could factor it too, but that wouldn't make much of a difference.

Answer 33

btw 0213, I doubt a test would want you to go that in depth or much even ask you to factor. cause like genius said it doesn't make a difference, they are just testing your algebra if they ask you to factor it