calculus
posted by Jamie .
Let f be a function defined by f(x)= arctan x/2 + arctan x. the value of f'(0) is?
It's 3/2 but I am not very clear on how to obtain the answer. I changed arctan x/2 into dy/dx=(42x)/(4sqrt(4+x^2)) but that's as far as I got. Could you please show me how to solve this problem? I would really appreciate it?
Start with this rule:
d/dx (arctan x)
= 1/(1 +x^2)
d/dx (arctan x/2) = (1/2)/[1 + (x/2)^2]
Evaluate them at x = 0 and add them. You will get 3/2.
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