- #1
syphonation
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Homework Statement
a) Find y' if y = x^2arcsinh(2x)
b) Find y' if y = xarctanh(x) + ln[(1-x^2)^1/2]
c) Find y' if y = arccoth[sqrt(x^2+4)]
Homework Equations
d/dx arcsinh(x) = 1/sqrt(1+x^2)
d/dx arctan(x) = 1/(1-x^2)
d/dx arccoth(x) = 1/(1-x^2)
The Attempt at a Solution
I am still having trouble mastering the Chain Rule. Trying these problems has made me second guess myself even more, so I have tried each one two different ways, both of which are apparently wrong. I'll write out what I have. I'm not necessarily expecting to get an answer, but I would appriceate it if someone could point out what I'm doing wrong.
a) y' = 2xarcsinh(2x) * [1/sqrt(1+2x^2)] * 2
y' = 2xarcsinh(2x) * [2/sqrt(1+2x^2)]
y' = (4xarcsinh(4x))/sqrt(1+2x^2)
OR
y' = 2x * [1/sqrt(1+2x^2)] * 2
y' = 4x/sqrt(1+2x^2)
b) y' = (1)arctanh(x) * [1/(1-x^2)] + [1/sqrt(1-x^2)] * [1/2(1-x^2)^-1/2] * (2x)
y' = [arctanh(x)/(1-x^2)] + [(x/sqrt(1-x^2))/sqrt(1-x^2)]
Then, multiply the top and bottom by sqrt(1-x^2) to get
y' = [arctanh(x)/(1-x^2)] + [x/(1-x^2)]
y' = [arctanh(x)+x]/(1-x^2)
OR, if I take y' of xarctanh(x) to be (1) * [1/(1-x^2)] = (1-x^2), I get
y' = (1+x)/(1-x^2)
c) y' = [1/1-sqrt(x^2+4)^2] * [1/2(x^2+4)^-1/2] * 2x
y' = [(x/sqrt(x^2+4))/1-x^2+4]
Then, multiply the top and bottom by sqrt(x^2+4) to get
y' = [x/(1-x^2+4)sqrt(x^2+4)]
y' = [x/(5-x^2)sqrt(x^2+4)]
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Sorry for this huge list. These problems are similar to each other, so most likely if I can figure out one, I will be able to get the rest. I've never excelled in math, so sorry if these are dumb questions..
If anyone has problems following what I did please ask me.
Thanks for your help.
Jimmy
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