Derivative formulas

Monday 30 January 2023

DERIVATIVE FORMULAS

Power Rule:

(d/dx) (xn ) = nxn-1

Derivative of a constant,

(d/dx) (a) = 0

Derivative of a constant multiplied with function f(x)

(d/dx) (a. f) = a (d/dx) f(x)

Sum Rule

(d/dx) [f (x)+ g(x)] = (d/dx)f (x)+ (d/dx) g(x)

(d/dx) [(f(x) - g(x)] = (d/dx)f(x) - (d/dx) g(x)

Product Rule

(d/dx) [f (x)*g(x)]= f(x) (d/dx)g(x) + g (x) (d/dx) f (x)

Quotient Rule

d/dx (f(x)/g(x))= g(x) d/dx f(x)-f (x)d/dx g(x)

g(x)²

Chain Rule

d/dx f[g(x)] = d/dx f[g(x)] d/dx g(x)

Derivatives of some trigonometric functions

(d/dx) sin x = cos x

(d/dx) cos x = -sin x

(d/dx) tan x = sec² x

(d/dx) cot x = - cosec² x

(d/dx) sec x = sec x tan x

(d/dx) cosec x = - cosec x cot x

Derivatives of elementary functions

(d/dx ) log x = 1/ x

(d/dx) e^{x =} e^x

(d/dx) a^x = a^x. log a , where a>0, a ≠ 1

_{e}

_{e}

(d/dx) √x =1/(2 √x)

(d/dx) log a^x = 1/(log a)x, x > 0

(d/dx) k =0 , where k is a constant

Derivatives of hyperbolic trigonometric functions

(d/dx ) sin hx = cos hx

(d/dx ) cos hx = sin hx

(d/dx ) tan hx = sech²x

(d/dx ) cot hx = -cosech²x

(d/dx ) sec hx = -sech x tanh x

(d/dx ) cosec hx = -cosech x cot hx

Derivatives of Inverse trigonometric functions

(d/dx ) sin ^-1 x = 1∕√(1-x²) , -1<x<1

(d/dx ) cos ^-1 x = -1∕√(1-x²) , -1<x<1

(d/dx ) tanh ^-1 x = 1/1+x²

(d/dx ) cot ^-1 x = -1/1+x²

(d/dx ) cosec ^-1 x = - 1/|x|√x²-1 , |x|>1

Derivatives of Inverse hyperbolic functions

(d/dx ) sinh ^-1 x = 1/(√x²+1)

(d/dx ) cosh ^-1 x = 1/(√x²-1)

(d/dx ) tanh ^-1 x = 1/(1-x²)

(d/dx ) coth ^-1 x = 1/x(√1-x²)

(d/dx ) cosech ^-1 x = 1/x(√1+x²)

Derivative Formula

Differentiation or derivative is a process, where we find the instantaneous

rate of change in function based on one of its variables. The most

common example is the rate change of displacement with respect to time,

called velocity. The opposite of finding a derivative is anti-differentiation.

Differentiation is a method of finding the derivative of a function

If x is a variable and y is another variable, then the rate of change of x with

respect to y is given by dy/dx.

This is the general expression of derivative of a function and is represented as

f'(x) = dy/dx, where y = f(x) is any function.

In real life, the derivatives are used by the people for maintaining a reservoir ,need

to know when will a reservoir overflow knowing the depth of the water at several

instances of time, Rocket scientists need to compute the precise velocity with

which the satellite needs to shot out from the rocket knowing the height of the

rocket at various times. Financial institutions needs to predict the changes in the

value of a particular stock knowing its present value. Many such cases it is

desirable to know how a particular parameter is changing with respect to some

other parameter.

PROBLEMS USING DERIVATIVE FORMULAS

For some constants a and b, find the derivative of

(i) (x − a) (x − b)

(ii) (ax2 + b)2

Solution:

(i) (x – a) (x – b)

f(x) = (x-a)(x-b)

f(x)= x²- (a+ b)x+ ab

By differentiating on both sides

d/dx f(x) = d/dx (x²- (a+ b)x+ ab)

=d/dx x²- (a +b)d/dx(x)+ d/dx (ab)

=2x - (a+ b) *1 + 0 [ since (d/dx) (xn ) = nxn-1

and (d/dx) (a) = 0 ]

= 2x -a-b

(ii) (ax2 + b)2

f(x) = (ax2 + b)2

Differentiating on both sides by

dy/dx f(x) = d/dx (ax2 + b)2

d/dx f(x) = a d/dx ( x2 )+ d/dx( b2 )

= 4ax (ax2 + b)

Find the derivative of

(i) (5x3 + 3x – 1) (x – 1)

(ii) x-3 (5 + 3x)

Solutions

(i) (5x3 + 3x – 1) (x – 1)

f(x) = (5x3 + 3x – 1) (x – 1)

Differentiating on both sides by product rule

d/dx f (x) = d/dx (5x3 + 3x – 1) (x – 1)

= (5x3 + 3x – 1) d/dx (x-1) +(x-1) d/dx (5x3 + 3x – 1)

= (5x3 + 3x – 1) d/dx (x) - d/dx (1) +(x-1) d/dx (5x3 ) + d/dx (3x )– d/dx (1)

= (5x3 + 3x – 1) * 1- 0 +(x-1)(15x2+3)

= 5x3 + 3x – 1+15x3+3x-15x2-3

= 20 x3 -15 x2+6x-4

(ii) x-3 (5 + 3x)

solution

f(x) = x-3 (5 + 3x)

Differentiating on both sides by product rule

d/dx f(x) = d/dx [x-3 (5 + 3x)]

= x-3 d/dx(5 + 3x)+(5 + 3x) d/dx x-3

= x-3 d/dx(5) + d/dx (3x)+(5 + 3x) d/dx x-3

= x-3 (0+3)+(5+3x) (-3 x-3-1)

= 3x-3-15x-4-9x-3

= -6 x-3-15x-4

Basic rules of exponents and powers

Tuesday 24 January 2023

Rules Of Exponents And Powers

1. am × an = am + n (Product rule)

2. am ÷ an = am–n (Quotient rule)

3. (am)n = am × n (Power of a power rule)

4. (ab)^m= a^mb^{m (Power of a product rule)}

5. (a/b)^m= a^m/b^{m (Power of a quotient rule)}

6. a^−m= 1/a^{m (Negative of an exponent)}

7. a⁰= 1 (zero exponent rule)

8. a¹ = a

EXPONENTS□

Exponents are used to write larger numbers into shorter form .

Such as 1000000= 10* 10*10*10*10*10 = 10⁶

The short notation 10 stands for the product of 10*10*10*10*10*10. Here 10 is

called the base and 6 is called the exponent. 10⁶ is read as 10 raise to the power of 6

or simply as sixth power of 10. 10⁶ is called the exponential form of

1000000.

Now, we have used numbers like 10,100,1000 etc.. while writing the numbers in an

expanded form.

eg ; 35421 =3 * 10000+ 5*1000+ 4*100+2*10+1

This can be written as 3* 10⁴+5*10³ +4* 10²+ 2* 10+1

However base can be another number also,

For example 64 = 2*2*2*2*2*2

= 2⁶

You can also extend this way when base is a negative number

For example (-27) = (-3)*(-3) *(-3)

= (-3)³

(-1)^{even number =} (1)

(-1)^{odd number}= (-1)

Examples;

(-2)⁵ = (-2)*(-2)*(-2)*(-2)*(-2)

(-5)⁴ = (-5)*(-5)*(-5)*(-5)

= 625

LAWS OF EXPONENTS

1. PRODUCT RULE ;

(Multiplying powers with the same base)

am × an = am + n

Examples

1. calculate a³×a⁵

Answer

By using the product rule

am × an = am + n

a³×a⁵ =a³⁺⁵

= a⁸

2. calculate 6^2×6⁴

Answer

By using the product rule

am × an = am + n

6²*6⁴= 6²⁺⁴

⁼6⁶

3.calculate a3×a4×a7

2
×a3×5a4×a3×5a4×a3×5a4

Answer

By using the product rule

am × an = am + n

a³×a4×a7 = a3+4+7

= a14

2. QUOTIENT RULE ;

( Dividing the powers with same base)

am ÷ an = am–n

Examples

1. calculate 8³ ÷8²

Answer

By using the quotient rule

am ÷ an = am–n

8³ ÷8² = 83-2

= 81

2. calculate c100 ÷ c90

Answer

By using the quotient rule

am ÷ an = am–n

c100 ÷ c90 = c100–90

= c10

3. calculate 911 ÷ 97

Answer

By using the quotient rule

am ÷ an = am–n

911 ÷ 97 = 911–7

= 94

3. POWER OF A POWER RULE

According to this law, if ‘a’ is the base, then the power raised to the power of

base ‘a’ gives the product of the powers raised to the base ‘a’, such as;

(am)n = am × n

Examples;

1. calculate (73)5

Answer

By using the power of a power rule

(am)n = am × n

(73)5 = 73 × 5

= 715

2.calculate (b2)6

Answer

By using the power of a power rule

(am)n = am × n

(b2)6 = b2 × 6

= b12

3.calculate (32)5 * (34)3

Answer

By using the power of a power rule

(am)n = am × n

(32)5 * (34)3 = 310 * 312

Now by using the product rule

am × an = am + n

310 * 312 = 310+12

= 322

4. POWER OF A PRODUCT RULE

According to this rule, for two or more different bases, if the power is same,

then;

(ab)^m= a^mb^m

Examples:

1. calculate 1/8 x 5-3

Answer

By using the power of a product rule

(ab)^m= a^mb^m

We can write, 1/8 = 2-3

2-3 x 5-3 = (2 × 5)-3 = 10-3

2.calculate 76 x 56

Answer

By using the power of a product rule

(ab)^m= a^mb^m

76 x 56 = (7 × 5)6

= 356

3. calculate b4 x c4

Answer

By using power of a product rule

(ab)^m= a^mb^m

b4 x c4 = (bc)⁴

^{5. POWER OF A QUOTIENT RULE}

According to this law, the fraction of two different bases with the same

power is

(a/b)^m= a^m/b^m

^Examples:

^{1. calculate}153/53

^Answer

^{By using the power of a quotient rule}

(a/b)^m = a^m/b^m

^153/53= (15/5)³

^{= (}3^{3) = 27}

^{2. calculate}104/1004

^Answer

^{By using the power of a quotient rule}

(a/b)^m = a^m/b^m

104/1004 = (10/100)⁴

⁼(1/10)⁴

3. calculate a⁷/b⁷

^Answer

^{By using the power of a quotient rule}

^{(a/b)^m = a^m/b^m}

a7/b7 = (a/b)⁷

⁼(a/b)⁷

6. NEGATIVE OF AN EXPONENT

According to this rule, if the exponent is negative, we can change

the exponent into positive by writing the same value in the denominator and the

numerator holds the value 1.

a^−m= 1/a^m

^Example:

^{Find the value of} 2-2

^Answer

^{By using the negative of an exponent rule}

a^−m= 1/a^m

2-2 ⁼1/2²

^=1/4

^{7. ZERO EXPONENT RULE}

According to this rule, when the power of any integer is zero, then its value is equal to 1,

a⁰= 1

Example:

Find the value of 7⁰

Answer

By using the zero exponent rule

a⁰= 1

7⁰ = 1

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