Question

prove the following derivative of hyperbolic function
1.d/dx(sinhx)=1/2(e^x-e^xd/dx(-x))
2.d/dx(coshx)=1/2(e^x
e^-x)-coshx

Answer 1

cosh(x) = (e^(x) + e^(-x))/2
sinh(x) = (e^(x) - e^(-x))/2
tanh(x) = (e^(x) - e^(-x)) / (e^(x) + e^(-x))

Split the fractions and differentiate as normal d/dx of e^(ux) = ue^(ux)

Answer 2

So for cosh(x):
d/dx[ (e^(x))/2 + e^(-x)/2 ]
(1/2)(e^x) + (-1/2)(e^-x)
Or, (e^(x) - e^(-x))/2 dx