Differentiation?

Let y = #y=(e^x+e^-x)/(e^x-e^-x)# Find #dy/dx# when# x=2#

1 Answer
Akaash P.
Dec 19, 2017

#-4/((e^2 - e^(-2))^2# or #-0.076.#

Explanation:

We must use quotient rule to evaluate this derivative. The way that I have remembered the specific application of quotient rule is through the saying " low d high - high d low over low low. " This may be confusing at first, but it will be handy in the long run. We will also be using the fact that the derivative of #e^x# is simply #e^x# and we will also be using chain rule in conjunction with this.

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

#=> (dy)/(dx) = ((e^x - e^(-x))(e^x - e^(-x)) - (e^x + e^(-x))(e^x + e^(-x)))/((e^x - e^(-x))(e^x - e^(-x)))#

Before we simplify, make sure you understand how our saying can be translated into the expression above. We multiplied the expression in the denominator with the derivative of the numerator (low d high) and subtracted that (minus) by the product of the expression in the numerator and the derivative of the expression in the denominator (high d low) . Finally, we put that all over the denominator squared (low low) .

#=> (dy)/(dx) = (e^(2x) - 2 + e^(-2x) - (e^(2x) + 2 + e^(-2x)))/((e^x - e^(-x))^2#

#=> (dy)/(dx) = -4/((e^x - e^(-x))^2#

So when #x = 2#, #(dy)/(dx)# is equal to #-4/((e^2 - e^(-2))^2#, which can be approximated as #-0.076.#

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