How do you find the derivative of $\frac{7 e^{x}}{2 e^{x} + 1}$ $(7e^x)/(2e^x+1)$ ?

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How do you find the derivative of $\frac{7 e^{x}}{2 e^{x} + 1}$ $(7e^x)/(2e^x+1)$ ?

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Answer 1 · 2016-06-04T10:15:10

$\frac{d}{d x} [\frac{7 e^{x}}{2 e^{x} + 1}] = \frac{7 e^{x} \cdot l n (e)}{2 e^{x} + 1} - \frac{14 e^{2 x} \cdot l n (e)}{{(2 e^{2} + 1)}^{2}}$ $d/(d x)[(7e^x)/(2e^x+1)]=(7 e^x*l n(e))/(2e^x+1)-(14e^(2x)*l n(e))/(2e^2+1)^2$

Explanation:

$\frac{d}{d x} [\frac{7 e^{x}}{2 e^{x} + 1}] = ?$ $d/(d x)[(7e^x)/(2e^x+1)]=?$

$((7e^x)^'*(2e^x+1)-(2e^x+1)^' *7e^x)/((2e^x+1)^2)$

$=[7e^x*l n (e)*(2e^x+1)- ( 2*e^x * l n (e)*7e^x)]/(2e^x+1)^2$

$d/(d x)[(7e^x)/(2e^x+1)]=(7 e^x*l n(e))/(2e^x+1)-(7e^x*2e^x*l n(e))/(2e^2+1)^2$

$d/(d x)[(7e^x)/(2e^x+1)]=(7 e^x*l n(e))/(2e^x+1)-(14e^(2x)*l n(e))/(2e^x+1)^2$

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How do you find the derivative of $\frac{7 e^{x}}{2 e^{x} + 1}$ $(7e^x)/(2e^x+1)$ ?

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How do you find the derivative of $\frac{7 e^{x}}{2 e^{x} + 1}$ $(7e^x)/(2e^x+1)$ ?