Question

How to find the derivative of (4x^4+8)/x^2?

Answer 1

Using the quotient rule, you have:
\[y=\frac{ 4x^4+8 }{ x^2 }=\frac{ u }{ v }\]
\[\text{So if:} \quad u=4x^4+8 \quad and \quad v=x^2\]
Then you need to calculate u' and v' \[\text {Then:} \frac{ dy }{ dx }=\frac{ vu'-uv' }{ v^2 }\]

Answer 2

so for the first section, vu', the derivative of u times v, and then for the other one it's the derivative of v times u.

Answer 3

yep exactly, if you work it out, I'll confirm your answer with you

Answer 4

Im doing something wrong :X

Answer 5

Okay I'll go through it with you:\[\text{If:}\quad u=4x^4+8 \quad \text{then what is u'? }\]
and the same for v'?

Answer 6

derivative of u is 16x^3 and v' is 2x

Answer 7

okay right so the formula is: \[\frac{ vu'-uv' }{ v^2 }=\frac{ (x^2)(16x^3)-(4x^4+8)(2x) }{ (x^2)^2 }\]\[=\frac{ 16x^5 - 8x^5 - 16x }{ x^4 }=\frac{ 8x^5-16x }{ x^4 }\] \[=\frac{ 8(x^5-2x) }{ x^4 }=\frac{ 8(x^4-2) }{ x^3 }\]

That's it

Answer 8

OOOO no wonder why, when i multiplied 2x*8 i only put 16

Answer 9

Do you follow it? :-)

Answer 10

Yup! Thanks! haha I forgot some of the rules

Answer 11

No problem!