Thursday, October 11, 2012 Names: M408N , , , . Worksheet #6 Take Home Worksheet. Work in groups of 3-­‐4 in class and turn it in to me on Monday!! 1. Complete the following derivative table and make a note reminding yourself how to derive it in case you forget it. Assume f and g are differentiable if applicable. Function Derivative sin(x) cos(x) cos(x) !sin(x) tan(x) sec 2 (x) csc(x) !csc(x)cot(x) sec(x) sec(x)tan(x) cot(x) !csc 2 (x) sin !1 (x) 1
1! x 2
cos!1 (x) sec!1 (x) cot !1 (x) 1! x 2
1
1+ x 2
tan !1 (x) csc!1 (x) 1
!
!
1
x x 2 !1
1
x x 2 !1
!
1
1+ x 2
e x e x ln(x) 1
x
x n nx n!1 f (g(x)) f '(g(x))! g'(x) f (x)
g(x)
g(x) f '(x) ! f (x)g'(x)
f (x)! g(x) f '(x)! g(x) + f (x)! g'(x) [ g(x)]
1. Recall that if f and g are inverses of each other, then g'(x) =
2
1
. f '(g(x))
Given that f (x) = x 5 + 2x 3 + x !1 find the equation of the tangent line to the graph of f !1 (x) at x = 3 . (Hint: Find f (1) ). Solution: d !1
1
f ) (3) =
.
(
dx
f '( f !1 (3)) Notice that f '(x) = 5x 4 + 6x 2 +1 and f (1) = 15 + 2(1)3 +1!1 = 3 " f !1 (3) = 1 So f '(x) = 5x 4 + 6x 2 +1 d !1
1
1
1
1
f ) (3) =
=
=
= (
!1
4
2
dx
f '( f (3)) f '(1) 5(1) + 6(1) +1 12
Since the derivative of the function at the point x=3 represents the slope of the tangent line at that point, then the equation of the line in point-­‐slope form is: y !1 =
1
(x ! 3) . 12
2. Find the derivative of f (x) =
tan !1 (x)" esin x
. (Don’t simplify!!! Just write the ln(x 2 + 3)
derivative.) Solution: !"ln(x 2 + 3)#$ % ! 1 % esin x + tan &1 x % esin x % cos x # & !"tan &1 (x)% esin x #$ % ! 1 % 2x #
'"1+ x 2
($
'" x 2 + 3
($
f '(x) =
2
!"ln(x 2 + 3)#$
3. Use multiple iterations of chain rule to compute the derivative of:
f (x) = 3 cos(4 ln(sin !1 (e 5x ))) . (Don’t simplify!!! Just write the derivative.) Solution: !2
$
"
$"
1"
4
1
!1
5x
$
( & "#e 5x $% & [ 5]
f '(x) = #cos(4 ln(sin (e )))% 3 & "#!sin(4 ln(sin !1 (e 5x )))$% & ' !1 5x ( & '
5x
2
3
# sin (e ) % '# 1! (e ) (%
4. If f (x) = sin(e 2 x!4 !1)" cos(csc!1 (x)) find f ' (2) . Solution: #
%
1
f '(x) = cos(e 2 x!4 !1)" #$e 2 x!4 " 2%& " cos(csc!1 (x)) + sin(e 2 x!4 !1)" '!sin(csc!1 (x))"
(
$
x x 2 !1 &
So, #
%
1
f '(2) = cos(e 2!2"4 "1)! #$e 2!2"4 ! 2%& ! cos(csc"1 (2)) + sin(e 2!2"4 "1)! '"sin(csc"1 (2))!
(
$
2 2 2 "1 &
)
,
"! %
1
= cos(0)! [ 2 ] ! cos $ ' + sin(0)! +(sin(csc(1 (2))!
.
#6&
*
2 2 2 (1 3
= 2!
= 3
2