Can anyone solve the following calculus function
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Can anyone solve the following calculus function

[From: ] [author: ] [Date: 11-06-06] [Hit: ]
Now use the Chain Rule to differentiate f(x) to find f(x).Are you OK with doing this?Hence f(0.3) = -0.3/[1- (0.f(0.......
Obtain f ’’ (0.3) forthe following:

(1)f(x)=arccosx

With clear explaination if you could please.

Thanks in advance

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This is a standard derivative. Use:

If f(x) = arccos(ax)

then f'(x) = - a/[sqrt(1-a^(2)x^(2) The domain for f(x) being -1
For f(x) = arccos(x), a is = 1. So we have:

f'(x) = 1/[sqrt(1-x^(2))] = (1-x^(2))^(-1/2)

Now use the Chain Rule to differentiate f'(x) to find f''(x). Are you OK with doing this? You will get:

f''(x) = -x /[(1-x^(2))^(3/2)]

Hence f''(0.3) = -0.3/[1- (0.3)^(2))^(3/2)]

f''(0.3) = -0.346 (3 sig.fig)
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