Obtain f ’’ (0.3) forthe following:
(1)f(x)=arccosx
With clear explaination if you could please.
Thanks in advance
(1)f(x)=arccosx
With clear explaination if you could please.
Thanks in advance
-
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)
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)