Let's differentiate both terms in the function f(x) = arctan(x/2) + arctan(x) separately and then add the results.
We'll start with arctan(x/2). The derivative of arctan(x/2) with respect to x is:
d/dx (arctan x/2) = (1/2) / [1 + (x/2)^2] = (1/2) / [1 + x^2/4].
Now, let's differentiate arctan(x) with respect to x:
d/dx (arctan x) = 1 / (1 + x^2).
To find f'(x), we need to add the derivatives of the two terms:
f'(x) = [(1/2) / (1 + x^2/4)] + [1 / (1 + x^2)].
Now, let's find f'(0) by substituting x = 0:
f'(0) = [(1/2) / (1 + 0)] + [1 / (1 + 0)] = (1/2) + 1 = 3/2.
So, f'(0) = 3/2.
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=(4-2x)/(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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