Student Note - Differentiation Techniques

8/12/2019 Student Note - Differentiation Techniques

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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


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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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

29


30/34

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

30


31/34

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

31


32/34

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


33/34

Similarly, the derivative of other inverse functions are given by

1. d

dxarccos x= 1

1 x2 , 1< x


34/34

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

Documents

Student Note - Differentiation Techniques