Higher Order Derivatives

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jexrox

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What is the best way to solve a complex function, when you need to find multiple derivatives? My equation is y=(x+2)^.5 * (x-3). My calculus book said that it is easier to solve these complicated functions if you take the natural log of both sides. This made my equations very messy. Is there a better way to easily and cleanly solve this?
 
jexrox said:
What is the best way to solve a complex function, when you need to find multiple derivatives? My equation is y=(x+2)^.5 * (x-3). My calculus book said that it is easier to solve these complicated functions if you take the natural log of both sides. This made my equations very messy. Is there a better way to easily and cleanly solve this?
Click to expand...

Which derivative did you need to find?

What is the "messy" result you got?
 


jexrox said:
What is the best way to solve a complex function, when you need to find multiple derivatives?

Solving equations and differentiating functions are not the same thing.


My equation is y = (x + 2)^(0.5) * (x - 3)

This function is not too complicated. It's a product. 8-)


My calculus book said that it is easier to solve these complicated functions if you take the natural log of both sides. This made my equations very messy. Is there a better way to easily and cleanly solve this?

Again, I'm not sure for what you're trying to "solve". Did they give you a value for either x or y? :?
Click to expand...

If you are trying to differentiate, instead (i.e., finding y', y'', et cetera), then you could begin by using the Product Rule. This will also require the Chain Rule when differentiating the sqrt(x + 2) part.

If you're trying to do something else, please provide us with the instructions that came with the exercise.

Cheers ~ Mark

 
The exact wording says find y', y" and y'" for the function

y = (x + 2)^(0.5) * (x - 3)

instead of using the product rule i did:
ln (y)= 1/2 ln (x+2)^.5 = ln (x-3)
then:

y'/y= 1/(2x+4) + 1/(x-3) --> y' = (x+2)^.5 * (x-3) * 1/(2x+4) + 1/(x-3)

However this will make it really complicated to get y". Does this make sense? Is it easier just to do the product rule and chain rule ?
 
jexrox said:
The exact wording says find y', y" and y'" for the function

y = (x + 2)^(0.5) * (x - 3)

instead of using the product rule i did:
ln (y)= 1/2 ln (x+2)^.5 = ln (x-3)
then:

y'/y= 1/(2x+4) + 1/(x-3) --> y' = (x+2)^.5 * (x-3) * 1/(2x+4) + 1/(x-3)

However this will make it really complicated to get y". Does this make sense? Is it easier just to do the product rule and chain rule ?
Click to expand...

[y"y -(y')[sup:2qww1ig1]2[/sup:2qww1ig1]]/y[sup:2qww1ig1]2[/sup:2qww1ig1] = -1/2 * 1/(x+2)[sup:2qww1ig1]2[/sup:2qww1ig1] - 1/(x-3)[sup:2qww1ig1]2[/sup:2qww1ig1]

y"/y = -1/2 * 1/(x+2)[sup:2qww1ig1]2[/sup:2qww1ig1] - 1/(x-3)[sup:2qww1ig1]2[/sup:2qww1ig1] + [1/(2x+4) + 1/(x-3)][sup:2qww1ig1]2[/sup:2qww1ig1] .... and so on....


"Simple" is in the eye of the beholder!!!
 
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