Week 7 – Derivative - II
c) x4 + y 4 = 2xy 3
d) cos3 x = sin(x + y)
e) x2/3 + y 2/3 = 5
Implicit Differentiation: An equation involving x and y may define y as a function of x.
This is called an implicit function.
For example,
2
2
x + y = 1,
f) (x2 + y 2 )2 = 50xy
Inverse Functions:
If f has an inverse, then
y
ye + 2x − cos y = 0
f f −1 (x) = x
define y implicitly.
y = x3 − 5x2 ,
y = cos(x2 − ex )
Differentiating both sides and using chain rule,
we obtain:
define y explicitly.
1
d −1
f (x) = 0 −1 dx
f f (x)
Exercise 7-1: We know that y is a function of
x. Is it explicit or implicit function?
Exercise 7-4: Using the information about
√
derivative of f = x2 , find the derivative of x.
a) y = x3 + sin x + xex
b) y =
1
1 + sin2 (πx)
Exercise 7-5: Using the information about
derivative of f = ex , find the derivative of ln x.
c) 3xy + x2 y 3 + x = 5
1
d
ln |x| = ,
dx
x
if a > 0 and a 6= 1
d) ex + ey = tan(xy)
The derivative of y can be found without solving
for y. This is called implicit differentiation. The
main idea is,
x 6= 0
d
1
loga x =
, x 6= 0
dx
x ln a
d x
a = ax ln a
dx
• Differentiate with respect to x
• Solve for y 0
Logarithmic Differentiation:
Logarithm transforms products into sums. This
helps in finding derivatives of some complicated
functions. For example if
Exercise 7-2: Find the slope of the tangent
line to the curves at the given points:
√
a) x2 + y 2 = 4 at (1, 3).
b) x2 + 3xy − (x − y)3 = 4
c) x4 + y 4 = 8xy
y=
(x3 + 1)(x2 − 1)
x8 + 6x4 + 1
at (1, 1).
then
at (2, 2).
ln y = ln(x3 + 1) + ln(x2 − 1) − ln(x8 + 6x4 + 1)
Exercise 7-3: Find
a)
dy
dx
y0
3x2
2x
8x7 + 24x3
= 3
+ 2
− 8
y
x + 1 x − 1 x + 6x4 + 1
√
√
x+ y =1
This is called logarithmic differentiation.
b) xy = tan xy
1
Exercise 7-6: Find y 0
x−1
a) y = ln
x+1
Function
Domain
Range
sin−1 x
−1 6 x 6 1
−π/2 6 y 6 π/2
cos−1 x
−1 6 x 6 1
06y6π
b) y = ln ln x
c) y = ex−ln x
tan−1 x
−∞ 6 x 6 +∞ −π/2 < y < π/2
d) y = 2−x
cot−1 x
−∞ 6 x 6 +∞
0<y<π
e) y = x sin ln x
sec−1 x
|x| > 1
0 6 y < π/2,
π/2 < y 6 π
csc−1 x
|x| > 1
−π/2 6 y < 0,
0 < y 6 π/2
f) y = xx
g) y = xln x
The graphs of some of these functions are given
below:
y
π/2
h) y = ln xln x
√
3
i) y =
x5 − x − 1
6x + 7
4
y = sin−1 x
x
−1
Inverse Trigonometric Functions:
We have to restrict the ranges to find the inverses of trigonometric functions. Otherwise,
they won’t be one-to one.
Note that if
x = tan y
1
−π/2
then
y
−1
y = arctan x or y = tan
x
π
y = cos−1 x
π/2
x
−1
2
1
y
y = tan−1 x
Function
π/2
Derivative
sin−1 x
√
1
1 − x2
cos−1 x
√
−1
1 − x2
x
−2
2
−π/2
Exercise 7-7: Find the following angles:
1
a) cos−1
2
1
b) cos−1 √
2
c) csc−1 2
1
1 + x2
cot−1 x
−1
1 + x2
sec−1 x
1
√
|x| x2 − 1
csc−1 x
−1
√
|x| x2 − 1
Exercise 7-9: Find y 0
a) y = cos−1 x3
−1
d) cot−1 √
3
b) y = sin−1 (1 − s)
We can find the derivatives of the inverse trigonometric functions as follows:
If y = arctan x then tan y = x. Evaluating the
derivative of both sides, we find
c) y = tan−1
y0 =
1
1+t
d) y = sec−1 ln x
(1 + tan2 y) y 0 = 1
⇒
tan−1 x
1
1
=
2
1 + tan y
1 + x2
e) y = csc−1
1
x
f) y = cos−1 sin x
Exercise 7-8: Find derivatives of y = arcsin x,
y = arccos x and y = sec−1 x using the same
method.
Review Exercises
Exercise 7-10: Find y 0
a) x2 y 2 + 8xy − 4x4 = 20
b) xey + yex + xy = 72
c) y 3 + y = 2 cos x
3
Exercise 7-11: Find y 0 at the indicated point:
a) x3 + y 3 = 9xy
at (2, 4)
b) 5x4/5 + 10y 6/5 = 15
at (1, 1)
Exercise 7-12: Find the equation of the tan√
gent line to x + xy = 6 at (4, 1)
Exercise 7-13: Find y 0
2 sin θ
a) y = ln
θeθ
√
b) y = s
s−1
c) y = 5x−3 ln x
d) y =
x−2
(3 + x)(1 + x2 )
x
e) y = arcsin(et )
f) y = arccos
1
1 − ln t
g) y = earctan x
h) y = sin sec−1 x
i) y = cos tan−1 x
j) y = ln3 (x2 + 1)
— End of WEEK —
Author: Dr. Emre Sermutlu
Last Update: November 9, 2016
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