That's odd - Solution
Graphs of Odd Functions with their Derivatives
y= t
From the image it appears that the derivative is an even function because if the derivative graph is reflected in the y axis it results in a copy of itself.
y=sin(t)
From the image it appears that the derivative is an even function.
y=t^3
y=t^5
y=sinh(t) using sinh(t)=0.5*(e^t-e^(-1*t))
The Maclaurin series for a general function, f(t), defined around t=0 and with all its derivatives defined at t=0 is given by:

If a function is odd then it will only have odd powers of t in its Maclaurin series, i.e. f(0) and f''(0) and f(n)(0) for n even, must all be 0.

Differentiating above we get:

We see that the power series for the derivative consists of only a constant term and even powers of t and therefore is even.
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