I got x^2 - 1/4x^2, so I found out the derivative of that and got x^4 + 1/16x^4. Why in the answer booklet it says, the first derivative is x^4 - 1/2 +1/16x^4 ?
Thank you
Update:Follow this series:
x^2 - 1/4x^2
1 + (y')^2 = 1+(x^4 - 1/2 +1/16x^4)
= x^4 + 1/2 + 1/16x^4
= (x^2 + 1/4x^2)^2
(Whats going on here)?
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Answers & Comments
Verified answer
What's going on is that you added 1 to (x^2 - (1/(4x^2))^2. This is why it's important that you added those steps
(x^2 - 1/(4x^2))^2 =>
x^4 - 2 * x^2 * 1/(4x^2) + 1/(16x^4) =>
x^4 - (2/4) * x^2/x^2 + 1/(16x^4) =>
x^4 - (1/2) * 1 + 1/(16x^4) =>
x^4 - (1/2) + 1/(16x^4)
At no point did anything say that the first derivative of x^4 - 1/(4x^2) is x^4 - 1/2 + 1/(16x^4). Words are important. Use them correctly.
So, when you add 1 to x^4 - 1/2 + 1/(16x^4), you get
x^4 + 1/2 + 1/(16x^4) =>
(x^2)^2 + 2 * x^2 * 1/(4x^2) + (1/(4x^2))^2 =>
(x^2 + 1/(4x^2))^2
Nothing happening here is anything more complicated than
(x + y)^2 = x^2 + 2xy + y^2
or
(x - y)^2 = x^2 - 2xy + y^2
I guess this may have been an arc-length question?
That would be the reason for calculating sqrt[1 + (y')^2],
so if y' were x^4 - 1/(4x)^2,
the sqrt[1 + (y')^2] would indeed come out to x^4 + 1/(4x)^2.