Finding the derivative using quotient rule?

Question

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

1485 views around the world

You can reuse this answer
Creative Commons License

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#