To differentiate the function, we'll have to calculate the
derivative of the expression under the logarithm, after that, we'll differentiate the
square root, and, at the end, we'll differentiate the expression under the square
root.
We'll note ln sqrt ( 17x^2 - cosx)=
u(v(t(x)))
Where u(v) = ln
v
v(t) = sqrt t
t(x) = 17x^2 -
cosx
Now, we'll differentiate t(x), with respect to
x:
t'(x) = (17x^2 -
cosx)'
t'(x) = (17x^2)' -
(cosx)'
t'(x) = 34x - (-sin
x)
t'(x) = 34x + sin
x
v(t(x))' = (sqrt t)' = t'(x)/2sqrt
t
v(t(x))' = (34x + sin x)/2sqrt (17x^2 -
cosx)
u'(v) = (ln
v)'
u'(v(t(x))) =
v(t(x))'/v
u'(v(t(x))) = (34x + sin x)/2sqrt (17x^2 -
cosx)*sqrt ( 17x^2 - cosx)
u'(v(t(x))) = (34x + sin
x)/2(17x^2 - cosx)
But f'(x) =
u'(v(t(x)))
f'(x) = (34x + sin x)/2(17x^2 -
cosx)
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