psi20 Posted October 15, 2004 Posted October 15, 2004 If you have f(x) = e^(-.25x + 2) is the derivative e^(-.25x + 2) ? If not, what is it?
psi20 Posted October 15, 2004 Author Posted October 15, 2004 No, I don't know anything except f'(x) = (f(x+h) - f(x))/h lim h--> 0 or something.
rakuenso Posted October 15, 2004 Posted October 15, 2004 edit f(x) = e^(-.25x + 2) f'(x) = e^(-.25x+2)*d/dx[-.025x+2] f'(x) = -.025x(e^(-.025x+2))
MandrakeRoot Posted October 15, 2004 Posted October 15, 2004 Simple rules for real valued differentiable functions : (f(x) g(x))' = f'(x) g(x) + f(x) g'(x) (f(x) + g(x))' = f'(x) + g'(x) [f(g(x))]' = f'(g(x))g'(x) Try showing their correctness with the definition of the derivative : Remember that a function f is differentiable at x if the following limit exists [math]\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}[/math] If so then the value is denoted f'(x) Mandrake
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