Course: Accelerated Engineering Calculus I
Instructor: Michael Medvinsky
10.8.3
Implicit Differentiation (3.5)
Some function are given in implicit form, for example one defines circle of radius 2 as
± 4 − x 2 , but it provides
x2 + y 2 =
4 . It is not so difficult to solve this function for y =
two different functions, which represent the upper and the lower half circle. In
additional, some functions are much more difficult to solve for y as a function of x, for
example the folium of Descartes x3 + y 3 =
6 xy or something more involved like
sin y 3 =
x 2 + cos xy 2 + y .
When we say that a function f(x) is implicitly defined by equation x3 + y 3 =
6 xy we
mean equation x3 + f ( x ) =
6 xf ( x ) is true for all values of x.
3
In order to differentiate implicate function we don’t need to formulate it explicitly, i.e.
we don’t need to solve it. Instead, we differentiate both sides of equation.
dy
of x 2 + y 2 =
4 we differentiate both sides to get
dx
2x
x
−
=
−
2x + 2 y ( x) y '( x) =
0 and then solve it to get y ' =
2y
y
Ex 2.
Ex 3.
To find
dy
3
of x3 + f ( x ) =
6 xf ( x ) we write 3 x 2 + 3 y 2 = 6 ( y + xy ') and solve it
dx
2
2
x + y = 2 y + 2 xy '
To find
for
x2 + y 2 − 2 y
2x
dy
To find of sin y 3 =
x 2 + cos xy 2 + y we write
dx
y' =
Ex 4.
3 y 2 y 'cos y 3 =2 x − ( y 2 + 2 xyy ') sin xy 2 + y '
3 y 2 y 'cos y 3 =
2 x − y 2 sin xy 2 − 2 xyy 'sin xy 2 + y '
(3 y
2
cos y 3 + 2 xy sin xy 2 − 1) y ' =2 x − y 2 sin xy 2
2 x − y 2 sin xy 2
y' = 2
3 y cos y 3 + 2 xy sin xy 2 − 1
10.8.3.1
Another method for implicit differentiation
There is another way to find derivative of implicit function. One consider implicit
function as function of several variables F ( x, y ) , so the equation like x 2 + y 2 =
25
Course: Accelerated Engineering Calculus I
Instructor: Michael Medvinsky
becomes F ( x, y ) = 25 , where F ( x, y=) x 2 + y 2 . In this case the derivative become
F
dy
x
2x
=
− x =
−
=
− , i.e. this provide the same result.
dx
Fy
y
2y
Ex 5.
Find dy/dx of x 2 y − y 2 x + xy − 5 =
0
2 xy − y 2 + y
F
dy
one can define F ( x, y ) = x 2 y − y 2 x + xy − 5 to get
=
− x =
− 2
dx
Fy
x − 2 yx + x
2 xy + x 2 y '− 2 yy ' x − y 2 + y ' x + y =
0
the other way is
− ( x 2 − 2 yx + x ) y ' ⇒ y ' =
−
2 xy − y 2 + y =
2 xy − y 2 + y
x 2 − 2 yx + x
10.8.3.2
Derivatives of complicated functions using implicit
differentiation
One can exploit the same idea to find derivatives of complicated functions like
Ex 1.
Ex 2.
1
1
y = x ⇒ y 2 =x ⇒ 2 yy ' =1 ⇒ y ' = =
2y 2 x
1
y =n x ⇒ y n =x ⇒ ny n −1 y ' =1 ⇒ y ' = n −1 =
ny
n
1
( x)
n
n −1
1
1
=
= 1−1/ n
−
n
1
nx
n ( x1/ n )
10.8.4 Derivatives of Inverse Trigonometric Functions (3.6)
One of the more interesting applications of implicit differentiation is differentiation of
inverse functions.
1
1
y =arcsin x ⇒ sin y =x ⇒ y 'cos y =1 ⇒ y ' =
= =
cos y cos arcsin x
y = arccos x ⇒ cos y = x ⇒ − y 'sin y = 1 ⇒ y ' = −
1
=
1 − sin 2 y
1
1
1
=−
=−
2
sin y
1 − cos y
1 − x2
1
1
y =arctan x ⇒ tan y =x ⇒ (1 + tan 2 y ) y ' =1 ⇒ y ' =
=
2
1 + tan y 1 + x 2
Note that all examples hint for the following Theorem:
1⇒
y=
f −1 ( x ) ⇔ f ( y ) =
x ⇒ f '( y ) y ' =
1
1
d −1
f ( x) =
y ' = = −1
dx
f '( y ) f '( f ( x ))
1
1 − x2
Course: Accelerated Engineering Calculus I
Instructor: Michael Medvinsky
10.8.5
Derivatives of Logarithmic Functions
We now find the derivative of logarithmic functions
1
1
y =log a x ⇒ a y =x ⇒ a y ( ln a ) y ' =1 ⇒ y ' = y
=
a ln a x ln a
For a special case a = e we get
Chain rule giveus:
Ex 1.
Ex 2.
d
1
ln x = .
dx
x
f '( x)
d
ln f ( x ) =
dx
f ( x)
d
1
d
2x +1
ln ( x 2 +=
x − 1)
⋅ ( x 2 +=
x − 1)
2
2
dx
x + x − 1 dx
x + x −1
1
d
ln cos ( sin x ) =
⋅ ( − sin ( sin x ) ⋅ cos x ) =− tan ( sin x ) cos x
cos ( sin x )
dx
The following is an interesting result
Ex 3.
x>0
d
d ln x
=
=
x ≠ 0, ln
| x|

dx
dx ln ( − x ) x < 0
1
 x

1
− 1 ( −=
1)
 x
x
x>0
1
=
x
x<0
10.8.6
Logarithmic Differentiation
Another trick to simplify differentiation involving products, quotients and powers:
=
y
Ex 4.
x5/3 x3 + 1
( 5 x + 3)
3
=
⇒ ln y ln
x5/3 x3 + 1
3
=
x + 3)
ln x5/3 x3 + 1 − ln ( 5=
3
( 5 x + 3)
5
1
=ln x5/3 + ln x 3 + 1 − 3ln ( 5 x + 3) = ln x + ln ( x3 + 1) − 3ln ( 5 x + 3)
3
2
2
 5 1 1 2x2
y ' 5 1 1 2x
5
5 
=
+
−
⋅
⇒
=
+
− 3⋅
y
3
'

 y=
3
3
y 3 x 2 x +1
5x + 3
5x + 3 
 3 x 2 x +1
 5 1 1 2x2
5  x5/3 x3 + 1
= 
+
−
⋅
3

3
5 x + 3  ( 5 x + 3 )3
 3 x 2 x +1
Ex 5.
y = x n ⇒| y |= | x |n ⇒ ln | y |= n ln | x |⇒
1
1
1
y'
= n ⇒ y ' = n y = n x n = nx n −1
y
x
x
x
Course: Accelerated Engineering Calculus I
Instructor: Michael Medvinsky
y ' ln x
1
=
+ x =
y 2 x
x
Ex 6.
ln x + 2
ln x + 2
ln x + 2 x
=
⇒=
y'
=
y
x
2 x
2 x
2 x
y= x
x
⇒ ln y = ln x
=
x
x ln x ⇒
Ex 7.
y= xsin x ⇒ ln y= sin x ln x ⇒
10.8.7
Ex 8.
The number e as a limit
Something more involved
ln lim x x
lim ln x x
y ' sin x
 sin x

=
+ cos x ln x ⇒ y '= 
+ cos x ln x  xsin x
y
x
 x

lim x ln x
x→0+
x→0+
=
e=
e x→0+
lim+ x x e=
x →0
lim+ x ln x =lim+
x →0
x →0
−y
y
y
y 1
ln x
1
=lim y =− lim y =0 ⇔ 0 ← y ≤ y ≤ 2 = → 0
y →∞ e
e
e
y
y
1/ x y →∞ e
lim x ln x
⇒ lim x x =
e x→0
x →0
=
e0 =
1
Consider now the derivative of f ( x ) = ln x at x=1:
' (1) lim
f=
x →0
1
ln (1 + x ) − ln1
f (1 + x ) − f (1)
1
x ) lim ln (1 + x ) x
= lim
= lim ln (1 +=
x →0
x →0 x
x →0
x
x
1
1
1
But we know that f ' (1)= = 1 thus we got lim ln (1 + x ) x =
1.
x →0
And we got another and very important limit of
1
e= e= e
1
lim ln (1+ x ) x
x→0
1


 1
lim 1 +  =  lim (1 + y ) y  = e 2
x →∞
 x
 y →0

 x + 2x −1 
2
Ex 10.
x →0
2
2x
Ex 9.
= lim e
Find lim  2

x →∞
 x −9 
 3 x2 +5 x −7 


x+4


1
1
= lim (1 + x ) x
ln (1+ x ) x
x →0
Course: Accelerated Engineering Calculus I
Instructor: Michael Medvinsky
2
2
3( x2 − 9)
3 x 2 + 5 x − 7 3 x 2 + 5 x − 7 − 20 + 20 3 ( x − 9 ) + 5 ( x + 4 ) 3 ( x − 9 )
=
=
=
=
+5
=
+5
x+4
x+4
x+4
x+4
x+4
2 ( x + 4)
x2 + 2x −1 x2 + 2x −1 − 8 + 8 x2 − 9 + 2x + 8
x+4
=
=
=1 + 2
=1 + 6 ⋅
=1 + 6 f ( x )
2
2
2
x −9
x −9
x −9
x −9
3( x2 − 9)
1 4
+
x+4
x x 2 = 0= 0 ⇒ lim 1 = ∞
lim f ( x )= lim
lim
=
x →∞
x →∞ 3 x 2 − 9
x →∞ f ( x )
(
) x→∞ 3 1 − 92  3
 x 
 x + 2x −1 
⇒ lim 

2
x →∞
 x −9 
2
= lim (1 + 6 y )
y →0
1
y
 3 x2 +5 x −7 


x+4


1
=
lim (1 + 6 f ( x ) ) f ( x )
f ( x )→0
+5
=
lim (1 + 6 y ) y lim (1 + 6 f ( x ) ) =
1
y →0
5
y →0
6
1


=  lim (1 + 6 y ) 6 y  = e6
y →0


Ex 1. Show that
 −1/6
e

f ( x) = 
1
 4 x  2 x +3
 2 x − 3 

x= −
3
2
3
2
is continuous at x = − .
else
1
1
1
 4 x  2 x +3
 2 x − 3 + 2 x + 3  2 x +3
 2 x + 3  2 x +3
lim 
lim
lim
=
=



1 +

x →−3/2 2 x − 3
x →−3/2
x →−3/2
2x − 3




 2x − 3 

y 
= lim 1 +

y =2 x + 3 y → 0
y−6

y −6 1
y y −6
lim e
=
y →0
1
y −6
e −1/6
=
1
+5
f ( x)