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Hello everyone, could someone explain to me where I am wrong because I can't find the mistake.

The quotient rule I am using is as follows\(f(x)={v(x) \over u(x)} \Rightarrow f´(x)={ v´(x)*u(x) - v(x)*u´(x)\over (u(x))^2}\)

with \(f(x)={2x\over 1-x^2}\) my result is \(f´(x)={2*(1-x^2)-2x*(-2x)\over (1-x^2)^2} \Rightarrow {2-6x^2 \over (1-x^2)^2}\)

However, the CAS calculators result is \(f´(x)={2\over 1-x^2}+{4*x^2\over(1-x^2)^2}\)

 Oct 21, 2019

Best Answer 

 #3
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\(\displaystyle \frac{2(1-x^{2}) +4x^{2}}{(1-x^{2})^{2}}=\)

 

Split into two terms.

 Oct 21, 2019
 #1
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-2x*(-2x) = +4x^2.

 Oct 21, 2019
 #2
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That explains a part of it but not entirely. 

How does  \({2-2x^2+4x^2\over (1-x^2)^2} \Rightarrow {2\over(1-x^2)} + {4x^2\over(1-x^2)^2}\) work?

Guest Oct 21, 2019
 #3
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Best Answer

\(\displaystyle \frac{2(1-x^{2}) +4x^{2}}{(1-x^{2})^{2}}=\)

 

Split into two terms.

Guest Oct 21, 2019

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