psi20 Posted October 15, 2004 Share Posted October 15, 2004 If you have f(x) = e^(-.25x + 2) is the derivative e^(-.25x + 2) ? If not, what is it? Link to comment Share on other sites More sharing options...
timo Posted October 15, 2004 Share Posted October 15, 2004 No it isn´t. Do you know the chain-rule? Link to comment Share on other sites More sharing options...
psi20 Posted October 15, 2004 Author Share Posted October 15, 2004 No, I don't know anything except f'(x) = (f(x+h) - f(x))/h lim h--> 0 or something. Link to comment Share on other sites More sharing options...
rakuenso Posted October 15, 2004 Share 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)) Link to comment Share on other sites More sharing options...
P_Rog Posted October 15, 2004 Share Posted October 15, 2004 d/dx[-.25x+2] = -.25 Link to comment Share on other sites More sharing options...
MandrakeRoot Posted October 15, 2004 Share 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 Link to comment Share on other sites More sharing options...
rakuenso Posted October 15, 2004 Share Posted October 15, 2004 d/dx[-.25x+2'] = -.25 oops =P Link to comment Share on other sites More sharing options...
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