Department of Mathematical Sciences
Instructor: Daiva Pucinskaite
Calculus I
February 23, 2016
Quiz 7
Derivatives of Tower Functions Find the derivative of the
function f given by
f (x) = xcos x
π
and evaluate the derivative at a = .
2
Recall:
1 0
(1) (ln(g(x)))0 =
g (x) here g(x) > 0,
g(x)
(2) ln(ab ) = b · ln a here a > 0,
(3) cos( π2 ) = 0 and sin( π2 ) = 1
f (x) = xcos x
(2)
=⇒
ln (f (x)) = ln (xcos x ) = cos x · ln x
=⇒
ln (f (x)) = cos x · ln x
=⇒
(ln (f (x)))0 = (cos x · ln x)0
| {z } |
{z
}
d
(ln(f (x)))
dx
(1)
Product Rule
d
(cos x·ln x)
dx
1
(f (x))0 = (cos x)0 · ln(x) + cos x · (ln(x))0
f (x)
1 0
1
=⇒
f (x) = − sin x · ln(x) + cos x ·
f (x)
x 1
·f (x)both sides
=⇒
f 0 (x) = f (x) − sin x · ln(x) + · cos x
x
1
f (x)=xcos x
0
cos x
=⇒
f (x) = x
− sin x · ln(x) + · cos x
x
π
π cos π2
π
π
1
0 π
f
=
− sin · ln + π · cos
2
2 2
2 2
2
0
π 2
(3) π
=
−1 · ln + · 0
2
2 π
π
= − ln
2
=⇒
Another version
Recall:
(1) eln a = a, a > 0
(2) ln(ab ) = b · ln a here a > 0,
(3) cos( π2 ) = 0 and sin( π2 ) = 1
0
(4) eg(x) = eg(x) g 0 (x)
(1)
f (x) = xcos x = eln x
f (x) = ecos x·ln x =⇒ f 0 (x)
=
(4)
=
Product
Rule
=
=
f
0
π 2
cos x
(2)
= ecos x·ln x
ecos x·ln x
0
ecos x·ln x · (cos x · ln x)0
x·ln x
0
0
e|cos{z
} · ((cos x) · ln(x) + cos x · (ln(x)) )
f (x) 1
cos x
x
· − sin x · ln(x) + cos x ·
x
π cos π2
π
π 1
π
=
− sin · ln + cos · π
2 2
2
2 2
0
π
2
π
(3)
=
−1 · ln + 0 ·
2
2
π
π
= − ln
2
2