Find dy/dx: y= 1/u^2 u=2x+3 I get 2x/(2x+3)^2
the easy form is mus include : -name of a drug -history of drug -street name -short +long term effect -how is it ingested -what effects does it have on body -anyother intersting facts
-4/(2x+1)^3
plz help me
i have to wirte an essay
I think you are in the wrong problem
what
i need to write my health easy idont want to fial
this is math
idc
can you go to the english section
i posted my question plz
you posted your question in my math question
plz
am sorry but can you help me
plz
if you go to "Home" at the top of the page and then click "English" I think, that is where you need to be
but theres no peole online yo help me i posted it 3x but no on helped me
I am a student just like you
To do these problems, I suggest utilize the chain rule as opposed to subbing in u into the original equation.
I got (2x+3)^-2 (2x) then i got my answer
y=1/u^2, rewrite as y=u^-2 to make it easier to differentiate. Using the chain rule. y'=-2u*u'. Find u' by differentiating u=2x+3 with respesct to x. Sub in u and u' into y'=-2u*u' and you have your solution.
plz
I really messed that one up. I don't know what I was doing. Now I get the answer: 1/(-2x-3)^3
you are missing a coefficienet of 4
ok I am going to take it step by step now: 1st step is to get u^-2 and 2
differentiate y to get y'=-2u^-3*u'. Substitute in u and u' and that's it.
yes then I get -2u^-3 (2) then -2(2x+3)^-3 (2) Am I good so far
yeah that's the right answer
ok then I get (-4x-6)^-3 (2) then I am wanting to make (-4x-6)^-3 into 2(-2x-3)^-3 This is where I think I am making the mistake
no you were already done. -2(2x+3)^-3 (2) is correct because the -2 multiples with the 2 to become -4. I don't really follow what you did in your last response
so will my answer be 2/-2(2x+3)^3
I feal like I need to do something with (2)
your -3 exponent is only applied to 2x+3 and not the -2, there is no need to bring it down
ok then it will be -4/(2x+3)^3
yes
Thank you for working through this problem with me and not just giving me the answer
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