Finding the derivative using quotient rule?

Question

$t(x)=(e^x+e^-x)/(e^x-e^-x)$

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Answer 1 · 2018-06-18T04:54:28

Given $t(x)=(e^x+e^-x)/(e^x-e^-x)$

Simplify a bit by multiplying by 1 in the form of $e^x/e^x$ :

$t(x)=e^x/e^x(e^x+e^-x)/(e^x-e^-x)$

$t(x)=(e^(2x)+1)/(e^(2x)-1)$

The quotient rule when $t(x)$ is of the form $(g(x))/(h(x))$ then:

$t'(x) = (g'(x)h(x)-g(x)h'(x))/(h(x))^2$

We observe that $g(x) = e^(2x)+1$ and $h(x) = e^(2x)-1$ , then:

$g'(x) = 2e^(2x)$ and $h'(x) = 2e^(2x)$

Substituting these values into the rule:

$t'(x) = ((2e^(2x))(e^(2x)-1)-(e^(2x)+1)(2e^(2x)))/(e^(2x)-1)^2$

Simplify the numerator:

$t'(x) = (-4e^(2x))/(e^(2x)-1)^2$