What is the derivative of #sinh(x)#?

Answer 1
#(d(sinh(x)))/dx = cosh(x)#
Proof: It is helpful to note that #sinh(x):=(e^x-e^-x)/2# and #cosh(x):=(e^x+e^-x)/2#. We can differentiate from here using either the quotient rule or the sum rule. I'll use the sum rule first: #sinh(x) = (e^x-e^-x)/2# #= (e^x)/2-(e^-x)/2# #=>(d(sinh(x)))/dx = d/dx((e^x)/2-(e^-x)/2)# #=d/dx(e^x/2)-d/dx(e^-x/2)# by the sum rule #=e^x/2-(-e^x/2)# by basic differentiation of exponential functions #=(e^x+e^-x)/2 = cosh(x).#
The quotient rule is just as easy: Let #u=e^x-e^-x# and #v=2# Hence #(du)/dx=e^x+e^-x# (by the sum rule and basic differentiation of exponential functions) and #(dv)/dx=0# Recalling that #(d(u/v))/dx = (v((du)/dx)-u((dv)/dx))/v^2# so #(d(sinh(x)))/dx=(2*(e^x+e^-x)-0)/2^2# #=(2*(e^x+e^-x))/4# #=(e^x+e^-x)/2=cosh(x).#
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Answer 2

The derivative of sinh(x) is cosh(x).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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