Question

How do you find the derivative of `(3x)/(2e^x+e^-x)`?

Answer 1

`=(6e^x(1-x) + 3e^(-x)(1+x) )/(2e^x + e^(-x))^2`

Explanation:

Differentiating this function is determined by applying the
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quotient rule.
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`d/dx(u/v)=(u'v-v'u)/v^2`
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`d/dx((3x)/(2e^x + e^(-x)))`
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`=((3x)'xx(2e^x + e^(-x))-(2e^x + e^(-x))'xx(3x))/(2e^x + e^(-x))^2`
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`=(3xx(2e^x + e^(-x))-(2e^x - e^(-x))xx(3x))/(2e^x + e^(-x))^2`
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`=((6e^x + 3e^(-x))-(6xe^x - 3xe^(-x)))/(2e^x + e^(-x))^2`
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`=(6e^x + 3e^(-x)-6xe^x + 3xe^(-x))/(2e^x + e^(-x))^2`
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`=(6e^x(1-x) + 3e^(-x)(1+x) )/(2e^x + e^(-x))^2`