Question

find the derivative:
y=(4xe^x)/(x^2+1)

Answer 1

u = 4xe^x
v = x^2 + 1

u' = 4(e^x + xe^x)
v' = 2x
(u'v -v'u) / v^2

(4(e^x+xe^x)*(x^2+1) - 2x*4xe^x) / (x^2 +1)^2

Answer 2

that seems very long to be the answer...

Answer 3

you have to solve the algebra

Answer 4

what algebra?

Answer 5

simplify it you see there are multiplications there

Answer 6

\[\ln(y)=\ln(\frac{4xe^{x}}{x^2+1})\]
\[\ln(y)=\ln(4xe^x)-\ln(x^2+1)\]
\[\ln(y)=\ln(4)+\ln(x)+\ln(e^x)-\ln(x^2+1)\]
\[\ln(y)=\ln(4)+\ln(x)+x \ln(e)-\ln(x^2+1)\]
\[\ln(y)=\ln(4)+\ln(x)+x-\ln(x^2+1)\]
\[\frac{y'}{y}=0+\frac{1}{x}+1-\frac{2x}{x^2+1}\]
\[y'=y(0+\frac{1}{x}+1-\frac{2x}{x^2+1})\]
\[y'=\frac{4xe^{x}}{x^2+1} \cdot(1+\frac{1}{x}-\frac{2x}{x^2+1})\]

Answer 7

is that the final answer?