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Find the equation for the line tangent to the curve at the given point?
 one year ago
 one year ago
Find the equation for the line tangent to the curve at the given point?
 one year ago
 one year ago

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swin2013Best ResponseYou've already chosen the best response.0
My work: y = x^3/2x + 1/2x ====> i split the fractions up dy/dx = x = 2/x^2 ====> this is where I messed up I think....
 one year ago

swin2013Best ResponseYou've already chosen the best response.0
it's not really difficult lol
 one year ago

ChlorophyllBest ResponseYou've already chosen the best response.0
No, it's not! It essentially need to be precisely though !
 one year ago

ChlorophyllBest ResponseYou've already chosen the best response.0
f'(x) = ( 2x³ 1 ) / 2x²
 one year ago

swin2013Best ResponseYou've already chosen the best response.0
well the numerator is not in parenthesis, does that make a difference?
 one year ago

swin2013Best ResponseYou've already chosen the best response.0
and even if you take a derivative for f(x) it wouldn't be 2x^31 right? it would be just 2x^3?
 one year ago

ChlorophyllBest ResponseYou've already chosen the best response.0
You need to use the equation editor if you don't know how to write properly!
 one year ago

swin2013Best ResponseYou've already chosen the best response.0
well either way, the derivative you have is wrong
 one year ago

swin2013Best ResponseYou've already chosen the best response.0
@zepdrix can you please help me? :)
 one year ago

ChlorophyllBest ResponseYou've already chosen the best response.0
@swin2013 I'm not wrong at this trivial level :P
 one year ago

swin2013Best ResponseYou've already chosen the best response.0
the derivative of x^31 is not 2x^31...
 one year ago

ChlorophyllBest ResponseYou've already chosen the best response.0
It proves that you have no clue about the formula !!!
 one year ago

swin2013Best ResponseYou've already chosen the best response.0
ooook. well i just quoted from what you said f'(x) was so yea
 one year ago

swin2013Best ResponseYou've already chosen the best response.0
I came here to search for help @Chlorophyll, so i can just wait on the next person to help me
 one year ago

ChlorophyllBest ResponseYou've already chosen the best response.0
You're poor soul !!!
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
Hey were you able to figure this one out ok? c:
 one year ago

swin2013Best ResponseYou've already chosen the best response.0
A little! sorry about the little mess earlier lolllll dy/dx = x  1/2x^2
 one year ago

swin2013Best ResponseYou've already chosen the best response.0
@zepdrix :) I'm not sure if you can still respond to this post since it's closed hahaha
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
Yes you can, I still get popups. It just doesn't show up on the main list.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
\[\large y=\frac{x^3+1}{2x}\qquad \rightarrow \qquad y=\frac{1}{2}x^2+\frac{1}{2}x^{2}\] \[\large y'=xx^{3}\]I think this is what you should be getting.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
Remember, when you subtract one from a NEGATIVE number, it actually gets LARGER in the negative direction.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
\[\large \left(x^{2}\right)\quad =\quad 2x^{21}\quad = \quad 2x^{3}\]Confused on this part?
 one year ago

swin2013Best ResponseYou've already chosen the best response.0
I get how you got 1/2 x^2 but not 1/2 x^2 i split up the fractions to kinda help me see it :/ so I got \[\frac{ x ^{3} }{ 2x } + \frac{ 1 }{ 2x }\] which is \[\frac{ 1 }{ 2 } x ^{2} + \frac{ 1 }{ 2 }x? \] :( sorry lolll
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
Oh i wrote it wrong? :C woops...\[\large y=\frac{ 1 }{ 2 } x ^{2} + \frac{ 1 }{ 2 }x\] \[\large y'=xx^{2}\]
 one year ago

swin2013Best ResponseYou've already chosen the best response.0
Yeeeee Yeah I got it :)) Thankss!
 one year ago

swin2013Best ResponseYou've already chosen the best response.0
well idk does that look right? y' = x  1/x^2?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
Yes. Understand how to proceed from here or no?
 one year ago

swin2013Best ResponseYou've already chosen the best response.0
Yes I do, thank youss!!!
 one year ago
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