+0

# Question about the the quotient rule for differential calculus

0
54
3

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

#3
+1

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

Split into two terms.

Oct 21, 2019

#1
0

-2x*(-2x) = +4x^2.

Oct 21, 2019
#2
0

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