derivative???

Question

what is the derivative of (4x^8-sqrt(x))/(8x^4)??
it gets really complicated and i dont know how far to go in simplifying!

we havnt learned derivatives of ln or log yet

Answer 1

So you want to find f'(x), when

f(x) = [4x^8 - sqrt(x)]/(8x^4)

This is clearly the usage of the quotient rule, but before we do that, let's change sqrt(x) to x^(1/2).

f(x) = [4x^8 - x^(1/2)]/[8x^4]

Proper mathematical etiquette desires the final to have (1) no negative exponents, (2) no complex fractions, (3) no radical (square root) in the denominator.

Quotient rule, verbal: "The derivative of the top times the bottom minus the derivative of the bottom times the top over the bottom squared."

At this point I'll be slowly calculating and grouping like terms.

f(x) = [4x^8 - x^(1/2)]/[8x^4]

f'(x) = ( [32x^7 - (1/2)x^(-1/2)][8x^4] - [32x^3][4x^8 - x^(1/2)] ) / [ (8x^4)^2 ]

f'(x) = [32x^7 - (1/2)(8)(x^(7/2))] - [32x^3][4x^8 - x^(1/2)] / [64 x^8]

f'(x) = [32x^7 - x^(7/2) - 128x^11 + 32x^(7/2)] / [64 x^8]

f'(x) = [32x^7 + 31x^(7/2) - 128x^11] / [64 x^8]

In the numerator, we can factor out x^(7/2), to obtain

f'(x) = x^(7/2) [32x^(7/2) + 31 - 128x^(15/2)] / [64 x^8]

And we can actually cancel the x^(7/2) with the denominator, which has x^8. This would change the denominator to x^(9/2).

f'(x) = [32x^(7/2) + 31 - 128x^(15/2)] / [64 x^(9/2)]

Since we have a radical on the bottom, we would have to multiply top and bottom by x^(1/2), giving us

f'(x) = [32x^4 + 31x^(1/2) - 128x^8] / [64 x^5]

And this is as far as I'm gonna go.

Just a piece of advice is that final exams in Calculus usually won't ask you to simplify further than you have to. They are only interested that you know HOW and have no intentions on reteaching something you learned in high school. However, it is good practice because solutions in the Calculus book are always usually in their most simplified form.

So just so you know, you don't have to worry too much about simplification.

Answer 2

try using log(In),.. if you are awared of that..

take In of both sides...that would make things lit bit easier//

-------------------------------------
just gonna do the logs way

y = (4x^8-sqrt(x))/(8x^4)
In y = In ((4x^8-sqrt(x))/(8x^4))

In y = In (4x^8-sqrt(x)) - In (8x^4)
= In (4x^8-sqrt(x)) - 4In (8x)

so
(1/y).dy/dx = ((32x^7 -0.5x^(-0.5))/(4x^8-sqrt(x))) - (4/x)

so
dy.dx = that all thing. y


d(In (y))/dx = (dy/dx) /y

see it's that easy... and i would also emphasize on the fact that no one really cares about simplifying unless..

Answer 3

Kitty Is this you? … :)!
http://www.osoq.com/funstuff/extra/extra01.asp?strName=Kitty