Differentiate $\frac{e^{8u}-e^{-8u}}{e^{8u}+e^{-8u}}$ ?

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Answer 1 · 2018-06-14T07:08:17

$f'(u)=(32*e^(16u))/(1+e^(16u))^2$

Multiplying numerator and denominator by

$e^(8u)$
we get
$f(u)=(e^(16u)-1)/(e^(16u)+1)$
now we get by the Quotient and chain rule

$f'(u)=((e^(16u)*16*(e^(16u)+1)-(e^(16u)-1)*e^(16u)*16))/(e^(16u)+1)^2$
simplifying we get the result above.

Answer 2 · 2018-06-14T09:18:19

Please see the explanation below.

We know that

$(e^(8u)-e^(-8u))/(e^(8u)+e^(8u))=tanh(8u)$

Therefore, the derivative is

$(tanh(8u))'=8sech^2(8u)$

$=8/cosh^2(8u)$

$=32/(e^(8u)+e^(-8u))^2$

$=(32e^(16u))/(e^(16u)+1)^2$

Differentiate \frac{e^{8u}-e^{-8u}}{e^{8u}+e^{-8u}}\frac{e^{8u}-e^{-8u}}{e^{8u}+e^{-8u}} ?