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Differentiation Techniques
Derivatives
Introduction:
Letfbe a functiony= f(x)
Let xrepresents a small change inx fromx0 tox0+ x, and the corresponding smallchange in y is
y= f(x0+ x) f(x0)then the average rate of change is
y
x=
change iny
change inx=
f(x0+ x) f(x0)x
is the average rate of change offbetween x0 and x0+ x.
x
y
f(x)
x
x0+ x
y
x0
f(x0+ x)
f(x0)
If we consider the average rate of change off over smaller and smaller intervals by
letting x approach 0,
dy
dx= lim
x0
y
x= lim
x0
f(x0+ x) f(x0)x
the limit of this average rate of change is called the instantaneous rate of change ofywith respect tox atx0.
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Sometimes we denote the derivative asy or f(x). Let x0=a and h = x, then
f(a) = limh0
f(a+h) f(a)h
(1)
This is to say that the slope of the tangent line touching the point on the curve rep-
resented by f(x) at x = a is f(a). This tangent line (red) is shown in the diagrambelow:
x
y
f(x)
h
a+h
f(a+h) f(a)
a
f(a+h)
f(a)
slope =f(a)
The slope of the blue line can be interpreted as the average rate of change off(x)
between aanda+h:
average =f(a+h) f(a)
h
Some examples of the instantaneous rate of change such as
tangent on the curve - change ofy with respect to the change ofx, velocity - rate of change of distance with respect to time, acceleration - rate of change of velocity with respect to time,
etc.
Example:
Given that f(x) =x3 + 1
(a) Find the average rate of change offwith respect to x over the interval [2, 6].
(b) Find the instantaneous rate of change offwith respect to x when x =2.
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Solution:
(a)
average =f(6) f(2)
6
2
=(6)3 + 1
(2)3 + 1
4 = 52(b)
f(2) = limh0
f(2 +h) f(2)h
= limh0
(2 +h)3 + 1 (2)3 + 1h
= limh0
h3 6h2 + 12h 8 + 1 + 7h
= limh
0
h3 6h2 + 12hh
= limh0
(h2 6h+ 12)= 12
Definition:
We write x= a+h and thus h= x a. As h approaches 0, this is equivalent to xapproachesa. By this setting, there is an equivalent way of stating the derivative of afunctionf at x= a:
f
(a) = limxa
f(x)
f(a)
x a (2)Example:
Find the derivative of the function f(x) = 2x2 4x+ 1 at x = 2.Solution:
By (1):
f(2) = limh0
f(2 +h) f(2)h
= limh
0
2(2 +h)2 4(2 +h) + 1
2(2)2 4(2) + 1h
= limh0
2(4 + 4h+h2) 8 4h+ 1 (8 8 + 1)h
= limh0
4h+ 2h2
h= lim
h0(4 + 2h)
= 4
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By (2):
f(2) = limx2
f(x) f(2)x 2
= limx2
2x2 4x+ 1 2(2)2 4(2) + 1
x 2= lim
x22x2 4x
x 2= lim
x22x(x 2)
x 2= lim
x2(2x)
= 4
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Example:
Iff(x) =
x 1, find f(x). Is there any difference between the domain off andf?Solution:
f(x) = limh0
f(x+h) f(x)h
= limh0
x+h 1 x 1
h
= limh0
x+h 1 x 1
h
x+h 1 + x 1x+h 1 + x 1
= limh0
x+h 1 (x 1)h(
x+h 1 + x 1)
= limh
0
h
h(
x+h
1 +
x
1)
= 1
x 1 + x 1 = 1
2
x 1Domain off(x) = [1, ), domain off(x) = (1, ).
Example:
Ifg(t) =2 t3 +t
, find g(t).
Solution:
g(t) = limh0 g(t+h) g(t)h
= limh0
2(t+h)3+(t+h)
2t3+t
h
= limh0
(2 t h)(3 +t) (2 t)(3 +t+h)h(3 +t+h)(3 +t)
= limh0
5hh(3 +t+h)(3 +t)
= limh0
5(3 +t+h)(3 +t)
= 5(3 +t)2
Example:
Iff(x) = 1
x+
x, find f(x).
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Solution:
f(x) = limh0
f(x+h) f(x)h
= limh0
1x+h+
x+h ( 1x +
x)
h= lim
h0
1x+h 1x
h +
x+h x
h
= limh0
x (x+h)hx(x+h)
+(
x+h x)(x+h+ x)h(
x+h+
x)
= limh0
hhx(x+h)
+ x+h x
h(
x+h+
x)
= 1x2
+ 1
2
x
Definition:
A function f is differentiable at a iff(a) exists.
Example:
Determine the domain where the function f(x) =|x| is differentiable.Solution:
The function fcan be written as
f(x) =|x|= x if x0x if x
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Case 1: Ifx >0,
f(x) = limh0
(x+h) xh
= 1
f(x) is differentiable for all x when x >0.
Case 2: Ifx
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Theorem:
Iff is differentiable at x= a, then f is continuous at x = a.
(Note: the opposite of the theorem is not always true, i.e., iff is continuous at x = athen f is differentiable at x = a. Weve already seen that the function f(x) =
|x
| is
continuous at 0 but not differentiable at 0.)
The above theorem also implies that iff is not continuous at x = a, then f is notdifferentiable atx= a. It is not true if we say that iffis not differentiable atx= a,thenf is not continuous atx = a, because there are functions not differentiable atx = a butis continuous at x= a. For instance, the function f(x) =|x| in the previous example.
Three possibilities of a function fails to be differentiable:
1. If the curve of the function has a sharp corner in it, there is no tangent at thispoint and the function is not differentiable at that point.
x
y
2. If the curve of the function has jump discontinuity, thenffails to be differentiable.
x
y
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3. If the curve of the functionfhas vertical tangent at x = a that is fis continuousat x = a however lim
xaf(x) =.
x
y
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Techniques of differentiation
1. A constant function:Iff(x) =c, where c is a constant, then f(x) = 0
e.g. Givenf(x) = 5, then f(x) = 0
2. Power Rule:Iff(x) =xn for n R, then f(x) =nxn1.e.g. Givenf(x) =x3, then f(x) = 3x2.
3. Coefficient Rule:
Ifc is a constant, then d
dx
cf(x)
= c
d
dxf(x)
e.g. d
dx(3x4) = 3
d
dxx4 = 3(4x3) = 12x3
4. Sum and Difference Rules:d
dx
f(x) g(x)= d
dxf(x) d
dxg(x)
e.g. Ify = 3x5 x4 + 4x2 7, findy.Solution:
y= 3(5x51) 4x41 + 4(2x21) 0 = 15x4 4x3 + 8x5. Product Rule:
Givenf(x) andg(x) are both differentiable, thend
dxf(x)g(x)
= f(x)
d
dxg(x)
+g(x)
d
dxf(x)
or ddx
f g= f g+fg
Proof:
d
dx
f(x)g(x)
= lim
h0
f(x+h)g(x+h) f(x)g(x)h
= limh0
f(x+h)g(x+h) f(x+h)g(x) +f(x+h)g(x) f(x)g(x)h
= limh0
f(x+h)
g(x+h) g(x)h
+g(x)f(x+h) f(x)
h
= limh0 f(x+h) limh0 g(x+h) g(x)h + limh0 g(x) limh0 f(x+h) f(x)h
= f(x)d
dx
g(x)
+g(x)
d
dx
f(x)
Example:
Iff(x) = (2x3 5)(x4 3x2), findf(x).
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Solution:
f(x) = (2x3 5)(4x3 + 6x3) + (6x2 + 0)(x4 3x2)= 8x6 + 12 20x3 30x3 + 24x6 18= 32x6 20x3 30x3 6
6. Quotient Rule:Iff and g are both differentiable, thend
dx
f(x)
g(x)
=
1
[g(x)]2
g(x)
d
dx
f(x)
f(x) ddx
g(x)
or d
dx
f
g
=
fg f gg2
Proof:
ddx
f(x)g(x)
= lim
h0
f(x+h)g(x+h)
f(x)
g(x)h
= limh0
f(x+h)g(x) f(x)g(x+h)
hg(x+h)g(x)
= limh0
f(x+h)g(x) f(x)g(x) +f(x)g(x) f(x)g(x+h)
hg(x+h)g(x)
= limh0
g(x)
f(x+h) f(x)h
f(x) g(x+h) g(x)h
g(x+h)g(x)
=
limh0
g(x) limh0
f(x+h) f(x)h
limh0
f(x) limh0
g(x+h) g(x)h
limh0
g(x+h) limh0
g(x)
= 1
[g(x)]2
g(x)
d
dx
f(x)
f(x) ddx
g(x)
Example:
Ify =x3 + 2x2 1
x 5 , findy.
Solution:
y = (x 5)(3x2 + 4x) (x3 + 2x2 1)(1)(x 5)2
= 3x3 + 4x2 15x2 20x x3 2x2 + 1
(x 5)2
= 2x3 13x2 20x+ 1
(x 5)2
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Exercise:
Findy if
(i) y= 6
t3
(ii) y= 7
x
(iii) y= x
(iv) y= 13
x2
(v) y=
x
(1 +x2)
(vi) y= (x2 + 5)(x7
3)
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Derivatives of Trigonometric functions
Iffis a function defined for all real numbers xby
f(x) = sin x
Here, sin x means the sine of the angle x whose measure in radian. There are othertrigonometric functions with similar convention such as cos, csc, sec, tan and cot. All thetrigonometric functions are continuous at every number in their domain.
As we know that f(x) is the slope of the tangent to the sine curve. From thedefinition of the derivative,
f(x) = limh0
f(x+h) f(x)h
= limh
0
sin(x+h) sin xh
= limh0
sin x cos h+ cos x sin h sin xh
= limh0
sin x
cos h 1
h
+ cos x
sin h
h
= limh0
sin x
limh0
cos h 1h
+ lim
h0cos x
limh0
sin h
h
Sincex is regarded as a constant when evaluating the limit as h0, we have
limh0
sin x= sin x
andlimh0
cos x= cos x
Here, the limit of limh0sin hh can be evaluated by drawing the table:
h sinhh
1 0.84147098480.1 0.9983341665
0.01 0.99998333340.001 0.9999998333
We see that sin hh 1 as h0. Thus,
limh0
sin h
h = 1
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And also that
limh0
cos h 1h
= limh0
cos h 1
h cos h+ 1
cos h+ 1
= limh0
cos2 h 1
h(cos h+ 1)
= limh0
sin2 hh(cos h+ 1)
= limh0
sin h
h sin h
cos h+ 1
= limh0
sin h
h lim
h0
sin h
cos h+ 1
= (1) 0
1 + 1
= 0
Therefore,
f(x) = ddx
(sin x) = cos x
x
y
x
y
32
32
2
2
2
2
32
32
1
-1
1
-1
y= f(x) = sin x
y= f(x) = cos x
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Example:
Differentiate y = x2 sin x.
Solution:
dy
dx =
d
dxx2 sin x= x2
d
dx(sin x) + sin x
d
dx(x2)
= x2 cos x+ 2x sin x
One also can prove that
1. d
dx(cos x) =sin x
2. d
dx(tan x) = sec2 x
3. d
dx(cot x) = csc2 x
4. d
dx(csc x) = csc x cot x
5. d
dx(sec x) = sec x tan x
Example:
Differentiate
f() = sin 1 + cos
Solution:
df
d =
cos (1 + cos ) sin ( sin )(1 + cos )2
= cos + cos2 + sin2
(1 + cos )2
= cos + 1
(1 + cos )2
= 11 + cos
Example:
Find the second order derivative of
g(t) = sec t
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Solution:
g(t) = sec t tan t
g(t) = sec t(sec2 t) + tan t(sec t tan t)
= sec3 t+ sec t tan2 t
Exercise:
1) f(x) = 4 cos x+ 2 sin x
2) f(x) =4x2 cos x
3) f(x) = sec x
1 + tan x
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The Chain Rule
Suppose y = f(u) =u2 and u = g(x) =
x+ 1. Sincey is a function ofu, and u isa function ofx,
y= f(u) =fg(x)= f(
x+ 1) = (
x+ 1)2
it follows that y is ultimately a function ofx. This procedure is called composition, i.e.,the new function is composed of two given functions f and g. The composite functionis defined by
(f g)(x) = fg(x)The domain off g is the set of all x in the domain ofg such that g is in the domainoff. The figure below shows the picture off g:
x
g(x)
f(g(x))
g f
f g
Note: In general, f g=g f. It is possible to take the composition of more functions,e.g. f g h= f{g[h(x)]}.
Ifg is differentiable atxandfis differentiable atg(x), then the composition functionf g is differentiable at x by
d
dx
f
g(x)
= (f g)(x) =f(g(x)) g(x)
Or ify = f(u) and u = g(x), then
dy
dx =
dy
dudu
dx
This is the Chain Rule.Note: For the composition of more functions, for example, f g h p= f(g{h[p(x)]}),the derivative offwith respect to x is
df
dx=
df
dg dg
dhdh
dp dp
dx
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Example 1:
Differentiate y =
2x3 + 3 by using the Chain Rule.
Solution:
Letu = 2x3 + 3 and thus y =
u= u1/2, we have
du
dx= 6x2 and
dy
du=
1
2u1/2
By using the Chain Rule,
dy
dx =
dy
dudu
dx
= 1
2u1/2 6x2
= 3x2
2x3 + 3
Example 2:
Differentiate y = cos(x3) by using the Chain Rule.
Solution:
Letu = x3 and thus y = cos u, we have
dudx
= 3x2 and dydu
=sin u
By using the Chain Rule,
dy
dx =
dy
dudu
dx= sin u 3x2= 3x2 sin x3
Example 3:
Differentiate y = cos3 xby using the Chain Rule.
Solution:
Letu = cos xand thus y = u3, we have
du
dx=sin x and dy
du= 3u2
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By using the Chain Rule,
dy
dx =
dy
dudu
dx= 3u2 ( sin x)=
3cos2 x sin x
Example 4:
Differentiate F(t) = (t2 + 3)100 by using the Chain Rule.
Solution:
Letu = t2 + 3 and thus F(u) =u100, we have
u= 2t and F(u) = 100u99
By using the Chain Rule,dy
dt = F u
= 100u99 2t= 200t(t2 + 3)99
Example 5:
Differentiate g () = (3 + 1)4 cos2 n by using the Product Rule and Chain Rule.
Solution:
The g() is a product of two functions, i.e., g = uv, if we let u = (3 + 1)4 andv= cos2 n, then by using the Product Rule,
dg
d =u
dv
d+v
du
d
Now, let p = 3 + 1 and u = p4, we have
dp
d = 32 and
du
dp = 4p3
By using the Chain Rule we havedu
d =
dp
d du
dp = 32 4p3 = 122(3 + 1)3
And also, let q= cos n and v = q2, we have
dq
d=n sin n and dv
dq = 2q
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By using the Chain Rule we have
dv
d =
dv
dq dq
d= 2q (n sin n) =2n cos n sin n
Combine these results we find that
dg
d = u
dv
d+v
du
d= (3 + 1)4(2n cos n sin n) + cos2 n 122(3 + 1)3= 2(3 + 1)3 cos n
62 cos n n(3 + 1) sin n
Exercise:
1. g(t) =
t 12t+ 1
6
2. y= tan(4x3 + 1)
3. f(t) =
t3 6
t
2
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Implicit DifferentiationSometimes we encountered the implicit function, e.g.:
(i) y3 +xy+ sin y = x
(ii) x3 +y3 =yx
(iii) ey + tan y+ 2x =x1
in which y is not expressed explicitly in terms ofx. The derivative of such an implicitfunction can be done by using the method of implicit differentiation. The implicitdifferentiation is the notion of a function defined implicitly and to determine derivativesby means of implicit differentiation.
The procedure to evaluate the derivative of the implicit function is as follows:
1. Differentiate both sides of the equation with respect to x.
2. Collect the terms with
dy
dx on one side of the equation.3. Factor out dydx .
4. Solve for dydx .
Suppose y = f(x), to differentiate the function F(y) with respect to x, i.e., dFdx
, weuse the Chain Rule as follow:
d
dx
F(y)
=
dy
dxdF
dy
Example:
Find dy
dx if
1) 2y= x2 + sin y
Solution:
d
dx(2y) =
d
dx(x2 + sin y)
dy
dx d
dy(2y) =
d
dx(x2) +
dy
dx d
dy(sin y)
2
dy
dx = 2x+ cos y dy
dx
2dy
dx cos y dy
dx = 2x
(2 cos y) dydx
= 2x
dy
dx =
2x
2 cos y
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2. x3 +y3 = 6xy
Solution:
d
dx(x3 +y3) =
d
dx(6xy)
3x2 +dy
dx d
dy(y3) = 6x
dy
dx+ 6y
3x2 + 3y2dy
dx = 6x
dy
dx+ 6y
3y2dy
dx 6x dy
dx = 6y 3x2
(3y2 6x) dydx
= 6y 3x2
dy
dx
= 2y x2
y2
2x
Exercise:
1.
x+
y= 4
2. x2y+xy2 = 3x
3. 4cos x sin y= 1
4.
xy= 1 +x2y
5. sin x+ cos y= sin x cos y
6. sin(x+y) =y2 cos x
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Derivatives of other functions
(I) Exponential function
A function of the form f(x) =bx,b >0, is called an exponential function with baseb, e.g. f(x) = 2x, g(x) = (1
3
)x, h(t) =t. The graph of exponential functions for variousbases ofb are shown below:
x
yy= 10x
y= 3x
y= 2xy= (0.1)x
1
The properties of the exponential functionf(x) =bx for b >0 and b= 1:
It is defined for all real values ofx, thus its domain is x(, ). It is continuous on the interval x(, ) and its range is f(x)(0, ). Ifb >1, the curve is increasing.
If 0< b
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Let u be a function ofx, i.e.,u= f(x). Then
d
dx
bu
= bu(ln b) dudx
Example:
Compute the derivatives off(x) = 4x3.
Solution:
Let y = x3, theny= 3x2. Thus
f(x) = d
dx4x
3
= dy
dx d
dy(4y)
= 3x2
4y ln 4
= 3x2 4x3 ln 4
Example:
Show that y = et +et/2 satisfies 2y y y= 0.
Solution:
y = et +et/2
y = et
1
2et/2
y = et +1
4et/2
then,
2y y y= 2et +
1
2et/2 et +1
2et/2 et et/2
= 0
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Example:
Findy of the implicit function ex2y =x+y.
Solution:
ddx(ex2
y) = ddx(x+y)
ex2y(x2y+ 2xy) = 1 +y
x2ex2yy+ 2xyex
2y = 1 +y
x2ex2yy y = 1 2xyex2y
(x2ex2y 1)y = 1 2xyex2y
y = 1 2xyex2y
x2ex2y 1
Exercise:
Find the derivatives of
1. g(t) = e3t2
2. f(x) =x2ex4
3. f() =esin2
cos
4. y= 1 + 2e3x
5. y=
1 +xe2x
6. y=ex
x
7. xey +yex = 1
8. y= cos(ex)
9. y= etanx
10. y= e2x cos x
11. y=ex +ex
ex ex
12. y= e3x
1 +ex
(II) Logarithmic function
A function of the form f(x) = logbx, forb >0, b= 1, x >0 is called the logarithmicfunction. Ifb= e, then logexln x, where we call ln the natural log.
Derivatives of log functions:
1. d
dx(logbx) =
1
x ln b
2. d
dx
logbu(x)
=
1
u(x) ln bdu
dx
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3. d
dx(ln x) =
1
x
4. d
dx
ln u(x)
=
1
u(x)
du
dx
Example:Find the derivatives of
1. y= ln(x2 1)Solution:
y= 1
x2 1d
dx(x2 1) = 2x
x2 1
2. y= ln
x
1 +x2
Solution:
y = 1
x/(1 +x2) d
dx
x
1 +x2
= 1 +x2
x
1 (1 +x2) x(2x)
(1 +x2)2
= 1 +x2
x
1 x2(1 +x2)2
= 1 x2x(1 +x2)
Exercise:
Find the derivatives of
1. y= log2
t
3
2. y= x2 log2(3 2x)3. y= ln(x2 + 10)
4. y= ln(cos )
5. log10
x
x 1
6. y= 5
ln x
7. y=
x ln x
8. y=1 + ln t
1
ln t
9. y= ln u
1 + ln(2u)
10. y= ln(sec x+ tan x)
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(III) Hyperbolic functions
The standard algebraic expressions for the hyperbolic functions defined by:
1. sinh x=ex ex
2
2. cosh x=ex +ex
2
3. tanh x=ex exex +ex
4. sechx= 2
ex
+ex
5. cschx= 2
ex ex
6. coth x=ex +ex
ex ex
Differentiation of the hyperbolic functions above are given by
1. d
dxsinh x= cosh x
Proof:
d
dxsinh x =
d
dx
ex ex
2
= ex +ex
2= cosh x
Similarly, it can be proved that
1. ddx
cosh x= sinh x
2. d
dxtanh x= sech2x
3. d
dxsechx= tanh x sechx
4. d
dxcschx= coth x cschx
5.
d
dxcoth x=csch2
x
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(IV) Logarithmic Differentiation technique
This technique is useful for differentiating functions that are complicated, composedof products, quotients and power.
Example:
Given y=x2 3
7x 14
(1 +x2)4 , find
dy
dx.
Solution:
By taking the logarithmic on both sides of the equation,
ln y = ln
x2 3
7x 14
(1 +x2)4
= ln x2 + ln( 3
7x 14) ln(1 +x2)4
= 2 ln x+
1
3ln 7x 14 4 ln(1 +x2
)
Now, differentiating both sides, we have
1
y
dy
dx = 2
1
x+
1
3
7
7x 14 4 2x
1 +x2
dy
dx =
2
x+
7
3(7x 14) 8x
1 +x2
y
=
2
x+
7
3(7x 14) 8x
1 +x2
x2 3
7x 14
(1 +x2)4
Example:
Given y=sin x cos x tan3 x
x , findy .
Solution:
By using the same technique,
ln y= lnsin x+ ln cos x+ 3 lntan x 12
ln x
and differentiate both sides yield
1
yy =
cos x
sin x+
( sin x)cos x
+3 sec2 x
tan x 1
2
1
x
y =
cos x
sin x (sin x)
cos x +
3 sec2 x
tan x 1
2x
y
=
cot x tan x+ 3
sin cos x 1
2x
sin x cos x tan3 x
x
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Example:
Given y= tt2
, findy.
Solution:
Taking the log on both sides yield
ln y= t2 ln t
and differentiate the equation above gives
1
yy = t2
1
t
+ 2t ln t
y = (t+ 2t ln t)y
= t(1 + 2 ln t)tt2
(V) Inverse Trigonometric function
Let y = sin x be defined in the domain 2 x
2 and thus the range of y is
(1, 1). In this domain the function is one-to-one. The inverse of sine is denoted byy = sin1 x or y = arcsin x. The curve represented by these functions are shown in thediagram below:
x
2
2
1 1
y
1
1
2
2
y= sin x
y= arcsin x
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In this diagram the blue line represents y = sin x and the red line represents y =arcsin x. The domain and the range of these curves are
y= sin x , y= arcsin xDomain: 2 x 2, 1x1Range: 1y1 ,
2 y
2
Note:
arcsin x= 1sin x
Let y = cos x be defined in the domain 0 x and thus has the range (1, 1),the inverse of cosine is denoted by y = cos1 x or y = arccos x. The curves representedby these functions are shown in the diagram below:
x2
1 1
y
1
1
2
y= cos x
y= arccos x
The domain and the range of these curves are
y= cos x , y= arccos xDomain: 0x , 1x1Range: 1y1 , 0y
Let y = tan x be defined in the domain 2 x
2, the inverse of the tangent
function is denoted byy = tan1 x or y = arctan x. The domain of the inverse tangent
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x
y
2
2
y= arctan x
function is the entire real line (all real values ofx) and the range is (2 , 2 ). The curveof the inverse tangent is shown in the figure in next page.
One question arose from these inverse functions is that how to differentiate the inversefunction? Let us address this question first by considering the derivative of arcsin x.Now, let y = arcsin x be defined in the domain1x1, then
sin y= x
which has the domain2 y 2 . The equation in above is an implicit function, andwe use the implicit differentiation to differentiate both sides with respect toxyields
d
dxsin y =
d
dxx
dydx
ddy
sin y = ddx
x
cos ydy
dx = 1
dy
dx =
1
cos y
= 1
1 sin2 y=
1
1
x2
Since y is defined in the domain x (1, 1), and we see that the curve has a positivegradient, therefore we only take the positive sign of the square root in above. We have
d
dxarcsin x=
11 x2
where1< x
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Similarly, the derivative of other inverse functions are given by
1. d
dxarccos x= 1
1 x2 , 1< x
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Example:
Findy ify = arccos(e2x).
Solution:
y = 1
1 (e2x)2 e2x 2
= 2e2x
1 e4x
Example:
Findy ify = arctan(cos x).
Solution:
y = 11 + (cos x)2
( sin x)
= sin x1 + cos2 x
Example:
Find dy
dx ify= arctan
x
a
+ ln
x ax+a
.
Solution: The differentiation is easier if we convert this function into
y = arctan
x
a
+ ln
x ax+a
= arctan
x
a
+
1
2ln(x a) 1
2ln(x+a)
Then,
dy
dx =
1
1 + ( xa)2
1
a
+
1
2
1
x a
12
1
x+a
= a
x2 +a2 + 1/2
x a 1/2
x+a
= a
x2 +a2+
a
x2 a2
= 2ax2
x4 a4