Introduction to calculus Calculus is a branch of ...

Differential Calculus by Dr. Syed Ibrahim

Introduction to calculus Calculus is a branch of mathematics involving or leading to calculations dealing with continuously varying functions. Calculus is a subject that falls into two parts: (i) differential calculus (or differentiation) and (ii) integral calculus (or integration). Differentiation is used in calculations involving velocity and acceleration, rates of change and maximum and minimum values of curves. Functional notation In an equation such as y = 3x2 + 2x − 5, y is said to be a function of x and may be written as y = f(x). An equation written in the form f(x)= 3x2 + 2x − 5 termed functional notation. The value of f(x)when x = 0 is denoted by f(0), and the value of f(x) when x = 2 is denoted by f(2) and so on. Thus when f(x)= 3x2 + 2x − 5, then f(0)= 3(0)2 + 2(0) − 5 = -5 and f(2)= 3(2)2 + 2(2) − 5 = 11 and so on. Problem 1. If f(x) = 4x2 - 3x + 2 find f(0), f(3), f(-1) an= f(3) – f(-1) f(x) = 4x2 - 3x + 2 0 2 f(0) = 4(0) – 3(0) + 2 = 2 f(3) = 4(3)2 – 3(3) + 2 = 36 _ 9 + 2 = 29 f(-1) = 4(-1)2 – 3(-1) + 2 =4+3+2=9 f(3)- f(-1)= 29 - 9 = 20 Problem 2. Given that f(x) = 5x2+ x – 7 determine: a)

f(2) ÷ f(1) b) f(3+a)

c)

f(3+a) - f(3)

1 CALCULUS by Dr. Ibrahim

d)

𝒇(𝟑+𝒂)− 𝒇(𝟑) 𝒂


f(x) = 5x2+ x – 7 f(2) = 5(2)2 + 2 - 7 = 15 f(1) = 5(1)2 + 1 - 7 = -1 15 f(2) ÷ f(1)= = −15

(i)

−1

(ii)

f(3+a)= 5(3+a)2 + (3+a) - 7 = 5(9+ a2 + 6a)+ (3+a)-7 = 45+ 5a2 +30a +3+a-7 = 5a2 +31a +41

f(3) = 5(3)2 + 3 - 7 = 45+3-7 = 41 2 f(3+a) - f(3)= 5a +31a +41- 41 = 5a2 +31a (iii)

(iv)

𝑓(3+𝑎)− 𝑓(3) 𝑎

=

5a2 +31a a

= 5a + 31

Problem 3. Diffrentiate from first principles f(x) = x2 and determine the value of the gradient of the curve at x =2. To diffrentiate from first principles means to find f’(x) by using the expression 𝑓(𝑥 + 𝑥) − 𝑓(𝑥) 𝑓 ′ (𝑥) = lim { } 𝑥→0 𝑥 𝑓(𝑥) = 𝑥 2 Substituting (x+x) for x gives 𝑓(𝑥 + 𝑥) = (𝑥 + 𝑥)2 = 𝑥 2 + 2𝑥𝑥 + 𝑥 2 , hence 𝑥 2 + 2𝑥𝑥 + 𝑥 2 − 𝑥 2 2𝑥𝑥 + 𝑥 2 ′ (𝑥) 𝑓 = lim { } = lim { } = lim {2𝑥 + 𝑥} 𝑥→0 𝑥→0 𝑥→0 𝑥 𝑥 As x0, [2𝑥 + 𝑥] [2𝑥 + 0]. Thus f’(x) = 2x, i.e., the differential coefficient of x2 is 2x. At x =2, the gradient of the curve, f’(x) = 2(2) = 4. Problem 4. Differentiate from first principles 2𝑓(𝑥) = 2𝑥 3 Solution: Substituting (x+x) for x gives 2𝑓(𝑥 + 𝑥) = 2(𝑥 + 𝑥)3 = 2(𝑥 + 𝑥)(𝑥 2 + 2𝑥𝑥 + 𝑥 2 ) = 2(𝑥 3 + 3𝑥 2 𝑥 + 3𝑥𝑥 2 + 𝑥 3 ) = 2𝑥 3 + 6𝑥 2 𝑥 + 6𝑥𝑥 2 + 2𝑥 3



𝑑𝑦 𝑓(𝑥 + 𝑥) − 𝑓(𝑥) = 𝑓 ′ (𝑥) = lim { } 𝑥→0 𝑑𝑥 𝑥 2𝑥 3 + 6𝑥 2 𝑥 + 6𝑥𝑥 2 + 2𝑥 3 − 2𝑥 3 = lim { } 𝑥→0 𝑥 6𝑥 2 𝑥 + 6𝑥𝑥 2 + 2𝑥 3 = lim { } 𝑥→0 𝑥 = lim {6𝑥2 + 6𝑥𝑥 + 2𝑥2 } Hence

𝑑𝑦 𝑑𝑥

𝑥→0

=𝑓

′ (𝑥)

= 6𝑥

2

Problem 5. Find the differential coefficient of 𝑦 = 4𝑥 2 + 5𝑥 − 3 and determine the gradient of the curve at x= -3 Solution:

𝑦 = 𝑓(𝑥) = 4𝑥 2 + 5𝑥 − 3 𝑓(𝑥 + 𝑥) = 4(𝑥 + 𝑥)2 + 5(𝑥 + 𝑥) − 3) = 4(𝑥 2 + 2𝑥𝑥 + 𝑥 2 ) + 5𝑥 + 5𝑥 − 3 = 4𝑥 2 + 8𝑥𝑥 + 4𝑥 2 + 5𝑥 + 5𝑥 − 3 𝑑𝑦 4𝑥 2 + 8𝑥𝑥 + 4𝑥 2 + 5𝑥 + 5𝑥 − 3 − (4𝑥 2 + 5𝑥 − 3) ′ (𝑥) =𝑓 = lim { } 𝑥→0 𝑑𝑥 𝑥 8𝑥𝑥 + 4𝑥 2 + 5𝑥 = lim { } 𝑥→0 𝑥 = lim {8𝑥 + 4𝑥 + 5} 𝑥→0

𝑑𝑦 = 𝑓 ′ (𝑥) = 8𝑥 + 5 𝑑𝑥 𝑑𝑦 At x = - 3, the gradient of the curve = 𝑓 ′ (𝑥) = 8(−3) + 5 = −19 𝑑𝑥 Problem 6. Using the general rule, differenciate the following with respect to x: 4 a)𝑦 = 5𝑥 7 𝑏) 𝑦 = 3√𝑥 𝑐)𝑦 = 2 𝑥 𝒅 (𝒂𝒙𝒏 ) = 𝒂𝒏𝒙𝒏−𝟏 𝒅𝒙 𝑑𝑦 𝑦 = 5𝑥 7 = 5 × 7 × 𝑥 7−1 = 𝟑𝟓𝒙𝟔 𝑑𝑥 𝑑𝑦 1 𝟑 𝑦 = 3√𝑥 = 3𝑥 1⁄2 = 3 × × 𝑥 1⁄2−1 = 𝒙−𝟏⁄𝟐 𝑑𝑥 2 𝟐



𝑦=

1 2 3 4 5 6 7 8 9 10 11 12 13

4 = 4𝑥 −2 𝑥2

𝑑𝑦 −𝟖 = 4 × (−2)𝑥 −2−1 = −8𝑥 −3 = 𝟑 𝑑𝑥 𝒙

Common Functions Constant Square Square root Exponential Logarithms Trigonometric functions

Inverse Trigonometric functions

Function(y) C X 𝑥2 √𝑥 𝑒𝑥 ln(x) sin(x) cos(x) tan(x) cot(x) sec(x) cosec(x) 𝑠𝑖𝑛−1 𝑥 −1

14

𝑐𝑜𝑠

15

𝑡𝑎𝑛−1 𝑥

16

𝑐𝑜𝑡 −1 𝑥

17

𝑠𝑒𝑐 −1 𝑥

18

−1

19 20 21 22 23 24 25

𝑐𝑜𝑠𝑒𝑐 Rules Multiplication by constant

Power Rule Sum Rule Difference Rule Product Rule Quotient Rule

𝑥

𝑥

Function Cf 𝑥𝑛 𝑢+𝑣 𝑢−𝑣 𝑢𝑣 𝑢 𝑣


Derivative(dy/dx) 0 1 2x 1⁄ 2√𝑥 𝑒𝑥 1⁄ 𝑥 cos(x) - sin(x) 𝑠𝑒𝑐 2 (𝑥) −𝑐𝑜𝑠𝑒𝑐 2 (𝑥) 𝑠𝑒𝑐𝑥 𝑡𝑎𝑛𝑥 −𝑐𝑜𝑠𝑒𝑐𝑥 𝑐𝑜𝑡𝑥 1 √1 − 𝑥 2 1 − √1 − 𝑥 2 1 1 + 𝑥2 1 − 1 + 𝑥2 1 𝑥√𝑥 2 − 1 1 − 𝑥√𝑥 2 − 1 Derivative 𝑐𝑓 ′ 𝑛𝑥 𝑛−1 𝑢′ + 𝑣 ′ 𝑢′ − 𝑣 ′ 𝑢𝑣 ′ + 𝑢′ 𝑣 𝑣𝑑𝑢 − 𝑢𝑑𝑣 𝑣2

Differential Calculus by Dr. Syed Ibrahim 2

4

5

𝑥3

Problem 7.Find the differential coefficient of 𝑦 = 𝑥 3 − Solution:

+ 4√𝑥 5 + 7

2 4 2 𝑦 = 𝑥 3 − 3 + 4√𝑥 5 + 7 = 𝑥 3 − 4𝑥 −3 + 4𝑥 5⁄2 + 7 5 𝑥 5 𝑑𝑦 2 5 = × 3𝑥 3−1 − 4(−3)𝑥 −3−1 + 4 ( ) 𝑥 (5⁄2)−1 + 0 𝑑𝑥 5 2 6 𝟔 𝟏𝟐 = 𝑥 2 + 12𝑥 −4 + 10𝑥 3⁄2 = 𝒙𝟐 + 𝟒 + 𝟏𝟎√𝒙𝟑 5 𝟓 𝒙 Problem 8. If 𝑓(𝑡) = 5𝑡 +

1

𝑓𝑖𝑛𝑑 𝑓 ′ (𝑡)

√𝑡 3

𝑓(𝑡) = 5𝑡 +

1 √𝑡 3

= 5𝑡 + 𝑡 −3⁄2

3 3 3 3 𝑓 ′ (𝑡) = 5(1)𝑡 1−1 + (− ) 𝑡 −3⁄2−1 = 5𝑡 0 − 𝑡 −5⁄2 = 5 − 5⁄2 = 5 − 2 2 2𝑡 2√𝑡 5 Power Rule 3 Problem 9. What is the derivative of x ? We can use the Power Rule, where n=3: 𝑑 𝑑𝑥

(𝑥 𝑛 ) = 𝑛𝑥 𝑛−1

𝑑 3 (𝑥 ) = 3𝑥 3−1 = 𝟑𝒙𝟐 𝑑𝑥 Problem 10. What is the derivative of (1/x)? We can use the Power Rule, where n= -1: 𝑑 1 𝑑 −1 −1 (𝑥 ) = −1𝑥 −1−1 = −𝑥 −2 = 2 ( )= 𝑑𝑥 𝑥 𝑑𝑥 𝑥 Product Rule Problem11. What is the derivative of cosx sinx? 𝑑(𝑢𝑣) = 𝑢𝑣 ′ + 𝑢′ 𝑣 Let u = cos(x) and v = sin(x) 𝑑 𝑑 𝑢′ = cos(x) = − sin(𝑥) 𝑎𝑛𝑑 𝑣′ = sin(x) = 𝑐𝑜 s(𝑥) 𝑑𝑥 𝑑𝑥 𝑑 {𝑐𝑜𝑠(𝑥)𝑠𝑖𝑛(𝑥)} = 𝑐𝑜𝑠(𝑥)𝑐𝑜𝑠(𝑥) − 𝑠𝑖𝑛(𝑥)𝑠𝑖𝑛(𝑥) = 𝒄𝒐𝒔𝟐 (𝒙) − 𝒔𝒊𝒏𝟐 (𝒙) 𝑑𝑥 1 + 𝑐𝑜𝑠2𝑥 1 − 𝑐𝑜𝑠2𝑥 1 + 𝑐𝑜𝑠2𝑥 − 1 + 𝑐𝑜𝑠2𝑥 2𝑐𝑜𝑠2𝑥 = −( = = 𝒄𝒐𝒔𝟐𝒙 )= 2 2 2 2 5 CALCULUS by Dr. Ibrahim


Chain Rule Problem 12. What is the derivative of 𝑠𝑖𝑛(𝑥 2 ) ? The Chain Rule says: the derivative of f(g(x)) = f'(g(x)) g'(x) Let u= x2 and y= sinu 𝑑𝑢 𝑑𝑥

= 2x

𝑑𝑦

and

= cosu

𝑑𝑢

So: 𝑑𝑦 𝑑𝑦 𝑑𝑢 = × = 𝑐𝑜𝑠(𝑢) × 2𝑥 = 2𝑥 𝑐𝑜𝑠(𝑥 2 ) 𝑑𝑥 𝑑𝑢 𝑑𝑥 Problem 13. What is the derivative of Let

u= sinx 𝑑𝑢 𝑑𝑥

= cosx

𝟏 𝐬𝐢𝐧(𝒙)

?

and

y=

1 𝑢

𝑑𝑦

and

𝑑𝑢

=−

1 𝑢2

𝑑𝑦 𝑑𝑦 𝑑𝑢 1 cos(𝑥) = × = − 2 × 𝑐𝑜𝑠𝑥 = − 𝑑𝑥 𝑑𝑢 𝑑𝑥 𝑢 𝑠𝑖𝑛2 (𝑥) Problem 14. What is the derivative of (𝟓𝒙 − 𝟐)𝟑 ? Let 𝑢 = 𝟓𝒙 − 𝟐 and y= 𝑢3 𝑑𝑢 𝑑𝑦 =5 and = 3𝑢2 𝑑𝑥

𝑑𝑢

𝑑𝑦 𝑑𝑦 𝑑𝑢 = × = 3𝑢2 × 5 = 3(5𝑥 − 2)2 × 5 = 𝟏𝟓(𝟓𝒙 − 𝟐)𝟐 𝑑𝑥 𝑑𝑢 𝑑𝑥 Problem 15. Find the derivative of y = 2 x3 - 4 x2 + 3 x – 5 𝑑𝑦 𝑑 𝑑 𝑑 𝑑 (2 𝑥 3 ) − (4 𝑥 2 ) + (3 𝑥) − (5) = 6𝑥 2 + 8𝑥 + 3 + 0 = 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥 = 𝟔𝒙𝟐 + 𝟖𝒙 + 𝟑 Problem 16. 𝐹𝑖𝑛𝑑 Let 𝑢 = 𝑥 2

𝑑𝑦 𝑑𝑥

, 𝑖𝑓 𝑦 = 𝑥 2 𝑒 𝑥 and

𝑣 = 𝑒𝑥

𝑑𝑦 𝑑𝑣 𝑑𝑢 =𝑢 + 𝑣 𝑑𝑥 𝑑𝑥 𝑑𝑥

𝑑𝑢 𝑑𝑥

= 2𝑥

and

𝑑𝑣 𝑑𝑥

= 𝑒𝑥

= 𝑥2. 𝑒𝑥 + 𝑒𝑥 . 2𝑥 = 𝒙𝐞𝒙(𝒙 + 𝟐)



Problem 17. 𝑊ℎ𝑎𝑡 𝑖𝑠

𝑑

𝑙𝑛𝑥

( 𝑑𝑥

𝑥

) 𝑢

By the quotient rule, 𝑑 ( ) = 𝑣

𝑣𝑑𝑢−𝑢𝑑𝑣 𝑣2

1 𝑥. − 𝑙𝑛𝑥. 1 1 − 𝑙𝑛𝑥 𝑑 𝑙𝑛𝑥 = ( )= 𝑥 2 𝑑𝑥 𝑥 𝑥 𝑥2 Problem 18. Find

𝑑𝑦 𝑑𝑥

, 𝑖𝑓 𝑦 = √𝑥 2 + 1 𝑦 = (𝑥 2 + 1)1⁄2 𝑑𝑦 1 2 𝑥 = (𝑥 + 1)1⁄2−1 . 2𝑥 = 𝑑𝑥 2 √𝑥 2 + 1

Problem 19. Find

𝑑𝑦 𝑑𝑥

, If y = e tan x 𝑑𝑦 𝑑𝑥

= 𝑒 𝑡𝑎𝑛𝑥 𝑠𝑒𝑐 2 𝑥

Problem 20. Find the derivative of y = 3x + sin(x) - 4cos(x) Solution

𝑑𝑦 𝑑 𝑑 𝑑 (3𝑥) + (𝑠𝑖𝑛𝑥) − 4 (𝑐𝑜𝑠 𝑥) = 3 + 𝑐𝑜𝑠 𝑥 − 4(−𝑠𝑖𝑛𝑥) = 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥 = 𝟑 + 𝒄𝒐𝒔 𝒙 + 𝟒𝒔𝒊𝒏𝒙 Problem 21. Find the derivative of y = 3ln(x) - 4ex 𝑑𝑦 𝑑 𝑑 𝟑 (4𝑒 𝑥 ) = − 𝟒𝒆𝒙 = 3𝑙𝑛(𝑥) − 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝒙 Problem 22. Find

Problem 23. Find

𝑑𝑦 𝑑𝑥

𝑑𝑦 𝑑𝑥

, , if y = sin(x2) 𝑑𝑦 = 2𝑥𝑐𝑜𝑠(𝑥 2 ) 𝑑𝑥 , if, y = sin 2 x

𝑦 = (𝑠𝑖𝑛𝑥)2

⇒

𝑑𝑦 = 2 sin 𝑥 cos 𝑥 = 𝐬𝐢𝐧 𝟐 𝒙 𝑑𝑥

𝑑𝑦

, if, y = ( x 3 + x − 1) 4 𝑑𝑦 𝑦 = (𝑥 3 + 𝑥 − 1)4 ⇒ = 4(𝑥 3 + 𝑥 − 1)3 . (3𝑥 2 + 1) 𝑑𝑥

Problem 24. Find

𝑑𝑥



Problem 25. What is the equation of the tangent line to the curve 𝑦 = 𝑒 𝑥 𝑙𝑛(𝑥) at the point (1, 0)? Solution:

The first step is to find the slope of the tangent line at x = 1, which is the value of the derivative of y at this point: slope at point (1,0) 𝑑𝑦 1 =| | = [𝑒 𝑥 . + 𝑙𝑛𝑥. 𝑒 𝑥 ] =𝑒 𝑑𝑥 𝑥=1 𝑥 𝑥=1 Since the point‐slope formula says that the straight line with slope m which passes through the point ( x 0, y 0) has the equation 𝑦 − 𝑦𝑜 = 𝑚(𝑥 − 𝑥𝑜 ) the equation of the desired tangent line is 𝑦 = 𝑒(𝑥 − 1) Problem 26. Consider the curve given implicitly by the equation 3𝑥 2 𝑦 − 𝑦 3 = 𝑥 + 1

What is the slope of this curve at the point where it crosses the x axis? Solution:

To find the slope of a curve defined implicitly (as is the case here), the technique of implicit differentiation is used: Differentiate both sides of the equation with respect to x; then solve the resulting equation for y′. 3𝑥 2 𝑦 − 𝑦 3 = 𝑥 + 1 3𝑥 2 𝑦 ′ + 6𝑥𝑦 − 3𝑦 2 𝑦 ′ = 1 𝑦 ′ (3𝑥 2 − 3𝑦 2 ) = 1 − 6𝑥𝑦 1 − 6𝑥𝑦 𝑦′ = 3(𝑥 2 − 𝑦 2 ) The curve crosses the x axis when y = 0, and the given equation clearly implies that x = − 1 at y = 0. From the expression directly above, the slope of the curve at the point (−1, 0) is |𝑦 ′ |(−1,0) = |

1 − 6𝑥𝑦 𝟏 = | 3(𝑥 2 − 𝑦 2 ) (−1,0) 𝟑

Exercise 𝑦 = 7𝑥 4 𝑦 = √𝑥 𝑦 = √𝑡 3 1 𝑦=6+ 3 𝑥

28𝑥 3 1⁄2√ 𝑥 3√ 𝑡 ⁄2 −3⁄𝑥 4


Differential Calculus by Dr. Syed Ibrahim 𝑦 = 3𝑥 −

1

+

1 𝑥

3+

√𝑥 5 1 𝑦= 2− +2 𝑥 √𝑥 7 𝑦 = 3(𝑡 − 2)2 𝑦 = (𝑥 + 1)3

1 2√𝑥 3

1 𝑥2 7

−

10 + 𝑥 3 2√𝑥 9 6𝑡 − 12 3𝑥 2 + 6𝑥 + 3 −

27. Differentiate 𝑓(𝑥) = 6𝑥 2 − 3𝑥 + 5 and find the gradient of the curve at a) x= -1, and b) x=2 12𝑥 − 3 𝑎) − 15

𝑏) 21

28. Find the differential coefficient of 𝑦 = 2𝑥 3 + 3𝑥 2 − 4𝑥 − 1 and determine the gradient of the curve at x = -2 6𝑥 2 + 6𝑥 − 4, 8 29. Determine the derivative of 𝑦 = −2𝑥 3 + 4𝑥 + 7 and determine the gradient of the curve at x = -1.5 −6𝑥 2 + 4, 29. 𝐼𝑓 𝑦 = 𝑐𝑜𝑠𝑎,

𝑑𝑦 𝑑

= −𝑎𝑠𝑖𝑛𝑎

30. 𝐼𝑓 𝑦 = cos(𝑎 + ),

𝑑𝑦 𝑑

− 9.5

𝑤ℎ𝑒𝑟𝑒 𝑎 𝑖𝑠 𝑎 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

= −𝑎𝑠𝑖𝑛(𝑎 + ) 𝑤ℎ𝑒𝑟𝑒 𝑎𝑎𝑛𝑑  𝑎𝑟𝑒 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑠

31. Differentiate the following with respect to the variable: 𝑎) 𝑦 = 2𝑠𝑖𝑛5

𝑎𝑛𝑑 𝑏) 𝑓(𝑡) = 3 𝑐𝑜𝑠2𝑡

𝑎) 𝑦 = 2𝑠𝑖𝑛5 𝑑𝑦 = 2 × 5 × 𝑐𝑜𝑠5 = 𝟏𝟎 𝒄𝒐𝒔𝟓 𝑑 𝑏) 𝑓(𝑡) = 3 𝑐𝑜𝑠2𝑡 𝑑𝑦 = 3(−2)𝑠𝑖𝑛2𝑡 = −𝟔 𝒔𝒊𝒏𝟐𝒕 𝑑 32. Find the differential coefficient of 𝑦 = 7 sin 2𝑥 − 3𝑐𝑜𝑠4𝑥



𝑦 = 7 sin 2𝑥 − 3𝑐𝑜𝑠4𝑥 𝑑𝑦 = (7 × 2) cos 2𝑥 − (3 × −4)𝑠𝑖𝑛4𝑥 = 14 cos 2𝑥 + 12 𝑠𝑖𝑛4𝑥 𝑑𝑥 33. Differentiate the following with respect to the variable: 𝑎) 𝑓() = 5sin(100𝜋 − 0.40) 𝑎)

𝐼𝑓

𝑎𝑛𝑑 𝑏) 𝑓(𝑡) = 2 cos(5𝑡 + 0.20)

𝑓() = 5sin(100𝜋 − 0.40)

𝑓 ′ () = 5[100𝜋 cos(100𝜋 − 0.40)] = 𝟓𝟎𝟎𝝅 𝐜𝐨𝐬(𝟏𝟎𝟎𝝅 − 𝟎. 𝟒𝟎) 𝑏)

𝐼𝑓

𝑓(𝑡) = 2 cos(5𝑡 + 0.20) 𝑓 ′ (𝑡) = 2 [−5sin(5𝑡 + 0.20)] = −𝟏𝟎 𝐬𝐢𝐧(𝟓𝒕 + 𝟎. 𝟐𝟎)

34. An alternating voltage is given by 𝑣 = 100 sin 200𝑡 𝑣𝑜𝑙𝑡𝑠, where t is the time in seconds. Calculate the rate of change of voltage when a) t= 0.005 s and b) t=0.01 s 35.

Differentiate with respect to x:

𝑎)

𝑦 = 4𝑠𝑖𝑛3𝑥 𝑎)

36.

12𝑐𝑜𝑠3𝑥

𝑏) 𝑦 = 2𝑐𝑜𝑠6𝑥 𝑏) 𝑦 = −12𝑠𝑖𝑛6𝑥

Given 𝑓() = 2𝑠𝑖𝑛3 − 5𝑐𝑜𝑠2, 𝑓𝑖𝑛𝑑 𝑓 ′ () 6𝑐𝑜𝑠3 + 10𝑠𝑖𝑛2

37. An alternating current is given by 𝑖 = 5 sin 100𝑡 𝑎𝑚𝑝𝑒𝑟𝑒𝑠, where t is the time in seconds. Determine the rate of change of current when a) t= 0.01 seconds [270.2 A/s] 38. If 𝑓(𝑡) = 3 sin(4𝑡 + 0.12) − 2 cos(3𝑡 − 0.72) 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒 𝑓 ′ (𝑡) [12 cos(4𝑡 + 0.12) + 6 sin(3𝑡 − 0.72)] 39. 𝐼𝑓 𝑦 = 𝑒 𝑎𝑥 , 𝑡ℎ𝑒𝑛

𝑑𝑦 = 𝑎𝑒 𝑎𝑥 𝑑𝑥

40. Differentiate the following with respect to the variable: 10 CALCULUS by Dr. Ibrahim


𝑎) 𝑦 = 3𝑒 2𝑥 𝑦 = 3𝑒 2𝑥 𝑓(𝑡) =

𝑎𝑛𝑑 𝑏) 𝑓(𝑡) = ⇒

4 3𝑒 5𝑡

𝑑𝑦 = 3(2𝑒 2𝑥 ) = 6𝑒 2𝑥 𝑑𝑥

4 4 −5𝑡 4 20 −5𝑡 (−5)𝑒 = 𝑒 = = − 3𝑒 5𝑡 3 3 3𝑒 5𝑡

41. 𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑡𝑒 𝑦 = 5𝑙𝑛3𝑥 𝑦 = 5𝑙𝑛3𝑥

𝑑𝑦 1 𝟓 =5×( )= 𝑑𝑥 𝑥 𝒙

42. Differentiate with respect to x: 𝑎) 𝑦 = 5𝑒 3𝑥

𝑏)

2 7𝑒 2𝑥

𝒂) 𝟏𝟓𝒆𝟑𝒙 43. Find the differential coefficient of 𝑎) 𝑦 = 12𝑥 3 𝐼𝑓 𝑦 = 𝑎𝑥 𝑛 𝑡ℎ𝑒𝑛

𝑑𝑦 𝑑𝑥

𝑇ℎ𝑢𝑠

b)

𝑦=

5 3

√𝑥 4

=

5 𝑥 4⁄3

𝑥3

𝑑𝑦 𝑑𝑥

= (12)(3)𝑥 3−1 = 𝟑𝟔𝒙𝟐

12 = 12𝑥 −3 , 𝑎 = 12 𝑎𝑛𝑑 𝑛 = −3 𝑥3

𝑑𝑦 𝟑𝟔 = (12)(−3)𝑥 −3−1 = −36𝑥 −4 = − 𝟒 𝑑𝑥 𝒙

44. Find the derivative of 𝑎) 𝑦 = 3√𝑥 𝑎) 𝑦 = 3√𝑥 = 3𝑥 1⁄2

12

= 𝑎𝑛𝑥 𝑛−1

a) 𝑆𝑖𝑛𝑐𝑒 𝑦 = 12𝑥 3 , 𝑎 = 12 𝑎𝑛𝑑 𝑛 = 3 𝑡ℎ𝑢𝑠 𝑏) 𝑦 =

𝑏) 𝑦 =

𝒃) −

𝑏) 𝑦 =

5 3

√𝑥 4

𝑑𝑦 1 3 𝟑 = 3 × × 𝑥 1⁄2−1 = 𝑥 −1⁄2 = 𝑑𝑥 2 2 𝟐√ 𝒙 = 5𝑥 −4⁄3

𝑑𝑦 −4 −20 −7⁄3 −20 −𝟐𝟎 = 5 × ( ) × 𝑥 −4⁄3−1 = 𝑥 = 7⁄3 = 3 𝑑𝑥 3 3 3𝑥 𝟑 √𝑥 7 11 CALCULUS by Dr. Ibrahim

𝟒 𝟕𝒆𝟐𝒙


45. Differentiate 𝑦 = 5𝑥 4 + 4𝑥 −

𝑦 = 5𝑥 4 + 4𝑥 −

1 2𝑥 2

+

1 √𝑥

− 3 with respect to x

1 1 1 −2 4 + − 3 = 5𝑥 + 4𝑥 − 𝑥 + 𝑥 −1⁄2 − 3 2 2𝑥 2 √𝑥

𝑑𝑦 1 1 = 20𝑥 3 + 4 + × 2𝑥 −2−1 − 𝑥−1⁄2−1 𝑑𝑥 2 2

𝑑𝑦 1 = 20𝑥 3 + 4 + 𝑥 −3 − 𝑥 −3⁄2 𝑑𝑥 2 𝑑𝑦 1 1 1 1 = 20𝑥 3 + 4 + 3 − 3⁄2 = 20𝑥 3 + 4 + 3 − 𝑑𝑥 𝑥 𝑥 2𝑥 2√𝑥 3 46. Find the differential coefficients of a) 𝑦 = 3 sin 4𝑥 and b) 𝑓(𝑡) = 2𝑐𝑜𝑠3𝑡 with respect to the variable

a) 𝑦 = 3 sin 4𝑥

;

𝑑𝑦 𝑑𝑥

= 3 × 4 cos 4𝑥 = 𝟏𝟐 𝐜𝐨𝐬 𝟒𝒙

; 𝑓 ′ (𝑡) = 2 × (−3)𝑠𝑖𝑛3𝑡 = −𝟔𝒔𝒊𝒏𝟑𝒕

b) 𝑓(𝑡) = 2𝑐𝑜𝑠3𝑡

47. Determine the derivative of a) 𝑦 = 3𝑒 5𝑥 b) 𝑓() =

2 𝑒 3𝜃

c) 𝑦 =

6 ln 2𝑥 𝑑𝑦

𝑦 = 3𝑒 5𝑥 ; 𝑓() = 𝑦 = 6 ln 2𝑥

𝑑𝑥 2

𝑒 3𝜃

;

= 3 × 5𝑒 5𝑥 = 𝟏𝟓𝒆𝟓𝒙

= 2𝑒 −3𝜃

;

𝑓′() = 2(−3)𝑒 −3𝜃 = −6𝑒 −3𝜃 =

−𝟔 𝒆𝟑𝜽

𝑑𝑦 𝟔 = 𝑑𝑥 𝒙

48. Determine the co-ordinates of the point on the graph 𝑦 = 3𝑥 2 − 7𝑥 + 2 where the gradient is -1. 12 CALCULUS by Dr. Ibrahim


Ans:

The gradient of the curve is given by the derivative 𝑦 = 3𝑥 2 − 7𝑥 + 2

6𝑥 − 7 = −1

;

𝑑𝑦 = 6𝑥 − 7 𝑡ℎ𝑒𝑛 𝑑𝑥

⇒ 6𝑥 = −1 + 7 = 6

⇒ 𝑥=𝟏

𝑦 = 3(1)2 − 7(1) + 2 = 3 − 7 + 2 = −𝟐

When x = 1,

Hence the gradient is -1, at the point (1, -2) 49. Find the gradient of the curve 𝑦 = 3𝑥 4 − 2𝑥 2 + 5𝑥 − 2 at the points (0,-2) and (1,4) 𝑦 = 3𝑥 4 − 2𝑥 2 + 5𝑥 − 2 𝑑𝑦 = 12𝑥 3 − 4𝑥 + 5 𝑑𝑥 At the point (0,-2), x = 0. Thus 𝑑𝑦 = 12(0)3 − 4(0) + 5 = 𝟓 𝑑𝑥 At the point (1,4), x = 1. Thus 𝑑𝑦 = 12(1)3 − 4(1) + 5 = 12 − 4 + 5 = 𝟏𝟑 𝑑𝑥 50. Differentiate

𝑎) 𝑦 = 5𝑥 5

;

𝑏) 𝑦 = 2.4𝑥 3.5 𝑐) 𝑦 =

1 𝑥

𝑑) 𝑦 =

−4 = −4𝑥 −2 2 𝑥

; ;

𝑑𝑦 𝑑𝑥

= 25𝑥 4

𝑑𝑦 = (2.4 × 3.5)𝑥 3.5−1 = 8.4𝑥 2.5 𝑑𝑥 𝑑𝑦 1 = − 2 𝑑𝑥 𝑥 ;

𝑑𝑦 8 = 8𝑥 −3 = 3 𝑑𝑥 𝑥



𝑒) 𝑦 = 2𝑥 𝑓) 𝑦 = 2√𝑥 3

𝑔) 𝑦 = 3 √𝑥 5 = 3𝑥 5/3 4

ℎ) 𝑦 = 𝑖) 𝑦 =

√𝑥 −3 3

√𝑥

= 4𝑥 −1/2

= −3𝑥 −1/3

;

;

𝑑𝑦 =2 𝑑𝑥

;

𝑑𝑦 1 1 = 2× = 𝑑𝑥 2 √𝑥 √𝑥

;

2 𝑑𝑦 5 5 3 = 3 × 𝑥 3−1 = 5𝑥 3 = 5 √𝑥 2 𝑑𝑥 3

;

𝑑𝑦 −1 − 1 −1 −2 =4× 𝑥 2 = −2𝑥 −3/2 = 𝑑𝑥 2 √𝑥 3

4 𝑑𝑦 −1 − 1 −1 −1 −1 = −3 × 𝑥 3 = −𝑥 − 3 = 4 = 3 𝑑𝑥 3 √𝑥 4 𝑥3

𝑗) 𝑦 = (𝑥 − 1)2

;

𝑑𝑦 = 2(𝑥 − 1) 𝑑𝑥

𝑘) 𝑦 = 2𝑠𝑖𝑛3𝑥

;

𝑑𝑦 = 2 × 3𝑐𝑜𝑠3𝑥 = 6𝑐𝑜𝑠3𝑥 𝑑𝑥

Differentiation of a product When y = u v, and u and v are both functions of x, then 𝑑𝑦 𝑑𝑣 𝑑𝑢 =𝑢 +𝑣 𝑑𝑥 𝑑𝑥 𝑑𝑥 This is known as product rule. 51. Find the differential coefficient of 𝑦 = 3𝑥 2 𝑠𝑖𝑛2𝑥 3𝑥 2 𝑠𝑖𝑛2𝑥 is a product of two terms. Let 𝑢 = 3𝑥 2 𝑎𝑛𝑑 𝑣 = 𝑠𝑖𝑛2𝑥 𝑑𝑦 𝑑𝑣 𝑑𝑢 =𝑢 +𝑣 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑦 = 3𝑥 2 (2𝑐𝑜𝑠2𝑥) + 𝑠𝑖𝑛2𝑥(6𝑥) = 6𝑥 2 𝑐𝑜𝑠2𝑥 + 6𝑥𝑠𝑖𝑛2𝑥 = 𝟔𝒙(𝒙𝒄𝒐𝒔𝟐𝒙 + 𝒔𝒊𝒏𝟐𝒙) 𝑑𝑥



52. Find the rate of change of y with respect to x : 𝑦 = 3√𝑥 𝑙𝑛2𝑥 Let 𝑢 = 3√𝑥 𝑎𝑛𝑑 𝑣 = 𝑙𝑛2𝑥 𝑑𝑦 𝑑𝑣 𝑑𝑢 =𝑢 +𝑣 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑦 1 3 3 3 𝟑 𝟏 = 3√𝑥 × + 𝑙𝑛2𝑥 = + 𝑙𝑛2𝑥 = (𝟏 + 𝒍𝒏𝟐𝒙) 𝑑𝑥 𝑥 𝟐 2 √𝑥 √𝑥 2 √𝑥 √𝒙 53. Differentiate 𝑦 = 𝑥 3 𝑐𝑜𝑠3𝑥𝑙𝑛𝑥 Let 𝑢 = 𝑥 3 𝑐𝑜𝑠3𝑥

𝑎𝑛𝑑 𝑣 = 𝑙𝑛𝑥

𝑑𝑦 𝑑𝑣 𝑑𝑢 =𝑢 +𝑣 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝒅𝒖 𝑑𝑣 1 = 𝑥 3 (−3𝑠𝑖𝑛3𝑥) + 𝑐𝑜𝑠3𝑥(3𝑥 2 ) = 𝒅𝒙 𝑑𝑥 𝑥 𝒅𝒚 1 = 𝑥 3 𝑐𝑜𝑠3𝑥 × ( ) + 𝑙𝑛𝑥 (−3𝑥 3 𝑠𝑖𝑛3𝑥 + 3𝑥 2 𝑐𝑜𝑠3𝑥) 𝒅𝒙 𝑥 = 𝑥 2 𝑐𝑜𝑠3𝑥 + 3𝑥 2 𝑙𝑛𝑥 (𝑐𝑜𝑠3𝑥 − 𝑥𝑠𝑖𝑛3𝑥) 𝑑𝑦 = 𝑥 2 (𝑐𝑜𝑠3𝑥 + 3𝑙𝑛𝑥 [𝑐𝑜𝑠3𝑥 − 𝑥𝑠𝑖𝑛3𝑥]) 𝑑𝑥 𝑑𝑦 𝑑𝑥 6𝑥 2 (𝑐𝑜𝑠3𝑥 − 𝑥𝑠𝑖𝑛3𝑥) 3 √𝑥 (1 + 𝑙𝑛 3𝑥) 2 3𝑡 (4𝑐𝑜𝑠4𝑡 𝑒 + 3𝑠𝑖𝑛4𝑡) 1 𝑒 4𝜃 ( + 4𝑙𝑛3)

HOME WORK 𝑦 2𝑥 𝑐𝑜𝑠3𝑥 √𝑥 3 𝑙𝑛 3𝑥 3

𝑒 3𝑡 𝑠𝑖𝑛4𝑡 𝑒 4𝜃 𝑙𝑛3



DIFFERENTIATION OF A QUOTIENT 𝑢 𝑣

When𝑦 = , and u and v are both functions of x, then 𝑑𝑢 𝑑𝑣 𝑑𝑦 𝑣 𝑑𝑥 − 𝑢 𝑑𝑥 = 𝑑𝑥 𝑣2 This is known as quotient rule. 15 CALCULUS by Dr. Ibrahim


54. Find the differential coefficient of 𝑦 = 𝑦=

4 sin 5𝑥 5𝑥 4

4 sin 5𝑥 5𝑥 4

𝑑𝑢 𝑑𝑣 𝑑𝑦 𝑣 𝑑𝑥 − 𝑢 𝑑𝑥 = 𝑑𝑥 𝑣2 Let 𝑢

= 4 𝑠𝑖𝑛5𝑥 𝑎𝑛𝑑 𝑣 = 5𝑥 4 𝑑𝑢 = 4 × 5𝑐𝑜𝑠5𝑥 = 20𝑐𝑜𝑠5𝑥 𝑑𝑥 𝑑𝑣 = 5 × 4𝑥 3 = 20𝑥 3 𝑑𝑥 𝑑𝑢 𝑑𝑣 𝑑𝑦 𝑣 𝑑𝑥 − 𝑢 𝑑𝑥 = 𝑑𝑥 𝑣2 𝑑𝑦 5𝑥 4 × 20𝑐𝑜𝑠5𝑥 − 4 𝑠𝑖𝑛5𝑥 × 20𝑥 3 = 𝑑𝑥 (5𝑥 4 )2 100𝑥 4 𝑐𝑜𝑠5𝑥 − 80𝑥 3 𝑠𝑖𝑛5𝑥 = 25𝑥 8 80𝑥 3 (5𝑥𝑐𝑜𝑠5𝑥 − 4 𝑠𝑖𝑛5𝑥) = 25𝑥 8 20𝑥 3 (5𝑥𝑐𝑜𝑠5𝑥 − 4 𝑠𝑖𝑛5𝑥) = 25𝑥 8 𝑑𝑦 4 = 5 (5𝑥𝑐𝑜𝑠5𝑥 − 4 𝑠𝑖𝑛5𝑥) 𝑑𝑥 5𝑥

55. Differentiate 𝑦 =

𝑡𝑒 2𝑡 2𝑐𝑜𝑠𝑡

Let 𝑢 = 𝑡𝑒 2𝑡 𝑎𝑛𝑑 𝑣 = 2𝑐𝑜𝑠𝑡

𝑑𝑢 𝑑𝑣 = 2𝑡 𝑒 2𝑡 + 𝑒 2𝑡 (1) 𝑎𝑛𝑑 = −2𝑠𝑖𝑛𝑡 𝑑𝑡 𝑑𝑡 16 CALCULUS by Dr. Ibrahim


𝑑𝑢 𝑑𝑣 𝑑𝑦 𝑣 𝑑𝑡 − 𝑢 𝑑𝑡 = 𝑑𝑡 𝑣2 𝑑𝑦 2𝑐𝑜𝑠𝑡(2𝑡 𝑒2𝑡 + 𝑒2𝑡 ) − 𝑡𝑒2𝑡 (−2𝑠𝑖𝑛𝑡) = 𝑑𝑡 (2𝑐𝑜𝑠𝑡)2

4𝑡 𝑒2𝑡 𝑐𝑜𝑠𝑡 + 2 𝑒2𝑡 𝑐𝑜𝑠𝑡 + 2𝑡𝑒2𝑡 𝑠𝑖𝑛𝑡 = 4𝑐𝑜𝑠2 𝑡 =

2 𝑒2𝑡 (2𝑡 𝑐𝑜𝑠𝑡 + 𝑐𝑜𝑠𝑡 + 𝑡𝑠𝑖𝑛𝑡) 4𝑐𝑜𝑠2 𝑡

𝑑𝑦 𝑒2𝑡 = (2𝑡 𝑐𝑜𝑠𝑡 + 𝑐𝑜𝑠𝑡 + 𝑡𝑠𝑖𝑛𝑡) 𝑑𝑡 2𝑐𝑜𝑠2 𝑡 56. Determine the gradient of the curve 𝑦 =

5𝑥 2𝑥 2 +4

at the point (√3,

Let 𝑢 = 5𝑥 𝑎𝑛𝑑 𝑣 = 2𝑥 2 + 4 𝑑𝑢 𝑑𝑣 = 5 𝑎𝑛𝑑 = 4𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑢 𝑑𝑣 𝑑𝑦 𝑣 𝑑𝑥 − 𝑢 𝑑𝑥 (2𝑥2 + 4)5 − 5𝑥 × 4𝑥 𝑑𝑥

At the point (√3,

=

=

𝑣2

√3 ) 2

2

(2𝑥2 + 4) 10𝑥2 + 20 − 20𝑥2 20 − 10𝑥2 = = (2𝑥2 + 4)2 (2𝑥2 + 4)2

√3 ), 2

𝑥 = √3

Hence the gradient 2

𝑑𝑦 20 − 10(√3) 20 − 30 10 𝟏 = = = − = − 2 2 𝑑𝑥 ( 6 + 4) 2 100 𝟏𝟎 [2(√3) + 4] 𝑑𝑦 𝑑𝑥

HOME WORK 𝑦 2𝑐𝑜𝑠3𝑥⁄ 𝑥3

−6 (𝑥𝑠𝑖𝑛3𝑥 + 𝑐𝑜𝑠3𝑥) 𝑥4 17 CALCULUS by Dr. Ibrahim


2𝑥 𝑥2 + 1

2(1 − 𝑥 2 ) (𝑥 2 + 1)2 3√(3𝑠𝑖𝑛2 − 4 𝑐𝑜𝑠2) 4𝑠𝑖𝑛2  1 (1 − 2 𝑙𝑛2𝑡) √𝑡 3 4𝑥 2𝑒 {(1 + 4𝑥)𝑠𝑖𝑛𝑥 − 𝑥𝑐𝑜𝑠𝑥} 𝑠𝑖𝑛2 𝑥 -18

3 √ 3 2𝑠𝑖𝑛2 𝑙𝑛2𝑡 √𝑡 2𝑥𝑒 4𝑥 𝑠𝑖𝑛𝑥 Find the gradient of the curve 𝑦 = 2𝑥 𝑎𝑡 𝑡ℎ𝑒 𝑝𝑜𝑖𝑛𝑡 (2, −4) 2 𝑥 −5

Function of a function 𝑑𝑦

𝑑𝑦

𝑑𝑢

If y is a function of x then = × 𝑑𝑥 𝑑𝑢 𝑑𝑥 This is known as ‘function of a function’ rule (or some times the chain rule) 57. For example 𝑦 = (3𝑥 − 1)9 then Let 𝑢 = 3𝑥 − 1, 𝑦 = 𝑢9 𝑑𝑢 =3 𝑑𝑥

58.

,

𝑑𝑦 = 9𝑢8 𝑑𝑢

𝑑𝑦 𝑑𝑦 𝑑𝑢 = × = 9𝑢8 × 3 = 27𝑢8 = 𝟐𝟕(𝟑𝒙 − 𝟏)𝟖 𝑑𝑥 𝑑𝑢 𝑑𝑥 𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑡𝑒 𝑦 = 3 cos(5𝑥 2 + 2) Let 𝑢 = 5𝑥 2 + 2 𝑡ℎ𝑒𝑛 𝑦 = 3 𝑐𝑜𝑠𝑢 𝑑𝑢 𝑑𝑦 = 10 𝑥 = −3𝑠𝑖𝑛𝑢 𝑑𝑥 𝑑𝑢 𝑑𝑦 𝑑𝑦 𝑑𝑢 = × = −3𝑠𝑖𝑛𝑢 × 10𝑥 = −30𝑥𝑠𝑖𝑛𝑢 𝑑𝑥 𝑑𝑢 𝑑𝑥 𝑑𝑦 = −𝟑𝟎𝒙𝒔𝒊𝒏(𝟓𝒙𝟐 + 𝟐) 𝑑𝑥

59. Find the derivative of: 𝑦 = (4𝑡 3 − 3𝑡)6 Let 𝑢 = 4𝑡 3 − 3𝑡, 𝑡ℎ𝑒𝑛 𝑦 = 𝑢6



𝐻𝑒𝑛𝑐𝑒

𝑑𝑢 𝑑𝑦 = 12𝑡 2 − 3 𝑎𝑛𝑑 = 6𝑢5 𝑑𝑡 𝑑𝑡

Using the function of a function rule 𝑑𝑦 𝑑𝑦 𝑑𝑢 = × = (6𝑢5 )(12𝑡 2 − 3) = 6(4𝑡 3 − 3𝑡)5 (12𝑡 2 − 3) 𝑑𝑡 𝑑𝑢 𝑑𝑡 = 𝟏𝟖(𝟒𝒕𝟐 − 𝟏)(𝟒𝒕𝟑 − 𝟑𝒕)𝟓 60. Determine the differential coefficient of: 𝑦 = √3𝑥 2 + 4𝑥 − 1 𝑦 = √3𝑥 2 + 4𝑥 − 1 = (3𝑥 2 + 4𝑥 − 1)1⁄2 𝐿𝑒𝑡 𝑢 = 3𝑥 2 + 4𝑥 − 1 ⇒ 𝐻𝑒𝑛𝑐𝑒

𝑦 = 𝑢1⁄2

𝑑𝑢 𝑑𝑦 1 −1⁄2 1 = 6𝑥 + 4 𝑎𝑛𝑑 = 𝑢 = 𝑑𝑥 𝑑𝑢 2 2 √𝑢

Using the function of a function rule 𝑑𝑦 𝑑𝑦 𝑑𝑢 1 3𝑥 + 2 𝟑𝒙 + 𝟐 (6𝑥 + 4 ) = = × = = 𝑑𝑥 𝑑𝑢 𝑑𝑥 2√𝑢 √𝑢 √𝟑𝒙𝟐 + 𝟒𝒙 − 𝟏 61. Differentiate 𝑦 = 3𝑡𝑎𝑛4 3𝑥 𝐿𝑒𝑡 𝑢 = 𝑡𝑎𝑛3𝑥 𝑡ℎ𝑒𝑛 𝑦 = 3𝑢4 𝑑𝑢 𝑑𝑦 = 3𝑠𝑒𝑐 2 3𝑥 𝑎𝑛𝑑 12𝑢3 𝑑𝑥 𝑑𝑢 𝑑𝑦 𝑑𝑦 𝑑𝑢 = × = 12𝑢3 × 3𝑠𝑒𝑐 2 3𝑥 = 12(𝑡𝑎𝑛3𝑥)3 × 3𝑠𝑒𝑐 2 3𝑥 𝑑𝑥 𝑑𝑢 𝑑𝑥 𝑑𝑦 = 36𝑡𝑎𝑛3 3𝑥𝑠𝑒𝑐 2 3𝑥 𝑑𝑥 2 62. Find the differential coefficient of 𝑦 = (2𝑡 3 − 5)4 2 𝑦= = 2(2𝑡 3 − 5)−4 3 4 (2𝑡 − 5) 𝑑𝑢 𝐿𝑒𝑡 𝑢 = 2𝑡 3 − 5 𝑡ℎ𝑒𝑛 𝑦 = 2(𝑢)−4 ⇒ = 6𝑡 2 𝑑𝑡 𝑑𝑦 −8 = −8𝑢−5 = 5 𝑑𝑢 𝑢 𝑑𝑦 𝑑𝑦 𝑑𝑢 −8 −48𝑡 2 2 = × = 5 × 6𝑡 = (2𝑡 3 − 5)5 𝑑𝑡 𝑑𝑢 𝑑𝑡 𝑢 19 CALCULUS by Dr. Ibrahim


1 2 3 4 5 6 7 8

𝑑𝑦 𝑑𝑥 2 5(6𝑥 − 5)(2𝑥 3 − 5𝑥)4 6𝑐𝑜𝑠(3 − 2) −10𝑐𝑜𝑠 4  𝑠𝑖𝑛 5(2 − 3𝑥 2 ) (𝑥 3 − 2𝑥 + 1)6 10𝑒 2𝑡+1 −20𝑡 𝑐𝑜𝑠𝑒𝑐 2 (5𝑡 2 + 3) 18𝑠𝑒𝑐 2 (3𝑦 + 1) 2𝑠𝑒𝑐 2  𝑒 𝑡𝑎𝑛

HOME WORK 𝑦 3 (2𝑥 − 5𝑥)5 2sin(3 − 2) 2𝑐𝑜𝑠 5  1 (𝑥 3 − 2𝑥 + 1)5 5𝑒 2𝑡+1 2cot(5𝑡 2 + 3) 6tan(3𝑦 + 1) 2𝑒 𝑡𝑎𝑛

Successive Differentiation When a function 𝑦 = 𝑓(𝑥) is differentiated with respect to x the differential 𝑑𝑦 coefficient is written as 𝑜𝑟 𝑓 ′ (𝑥). If the expression is differentiated again, the 𝑑𝑥

second differential coefficient is obtained and is written as

𝑑2𝑦

𝑑𝑥 2 𝑑3𝑦

𝑜𝑟𝑓 ′ ′(𝑥). By

successive differentiation further higher derivative such as 3 and 𝑑𝑥 obtained. 63. Thus if 𝑦 = 3𝑥 4 𝑑𝑦 𝑑𝑥

𝐼𝑓

= 12𝑥 3 ,

𝑑2 𝑦 𝑑𝑥 2

= 36𝑥 2 ,


= 72𝑥,

𝑓(𝑥) = 2𝑥 5 − 4𝑥 3 + 3𝑥 − 5,


𝑓𝑖𝑛𝑑

= 72 𝑎𝑛𝑑


𝑓 ′′ (𝑥)

𝑓(𝑥) = 2𝑥 5 − 4𝑥 3 + 3𝑥 − 5 𝑓 ′ (𝑥) = 10𝑥 4 − 12𝑥 2 + 3 ′(𝑥) 𝑓′ = 40𝑥 3 − 24𝑥 = 4𝑥(10𝑥 2 − 6)


=0

𝑑4𝑦 𝑑𝑥 4

may


64. Given 𝑦 = 2𝑥𝑒

−3𝑥

𝑦 = 2𝑥𝑒 −3𝑥

𝑑 𝑦 𝑑𝑦 𝑠ℎ𝑜𝑤 𝑡ℎ𝑎𝑡 2 + 6 + 9𝑦 = 0 𝑑𝑥 𝑑𝑥

⇒

𝑑𝑦 = 2𝑥(−3𝑒 −3𝑥 ) + 𝑒 −3𝑥 (2) = −6𝑥𝑒 −3𝑥 + 2𝑒 −3𝑥 𝑑𝑥

𝑑2 𝑦 = −6𝑥(−3𝑒−3𝑥 ) + 𝑒−3𝑥 (−6) + (−6𝑒−3𝑥 ) = 18𝑥𝑒−3𝑥 − 6𝑒−3𝑥 − 6𝑒−3𝑥 2 𝑑𝑥 𝑑2 𝑦 = 18𝑥𝑒−3𝑥 − 12𝑒−3𝑥 2 𝑑𝑥 𝑑2 𝑦 𝑑𝑦 𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑔 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒𝑠 𝑖𝑛𝑡𝑜 2 + 6 + 9𝑦 = 0 𝑔𝑖𝑣𝑒𝑠: 𝑑𝑥 𝑑𝑥 18𝑥𝑒 −3𝑥 − 12𝑒 −3𝑥 + 6(−6𝑥𝑒 −3𝑥 + 2𝑒 −3𝑥 ) + 9(2𝑥𝑒 −3𝑥 ) = 𝟏𝟖𝒙𝒆−𝟑𝒙 − 𝟏𝟐𝒆−𝟑𝒙 − 36𝑥𝑒 −3𝑥 + 𝟏𝟐𝒆−𝟑𝒙 + 𝟏𝟖𝒙𝒆−𝟑𝒙 = 𝟎 2

𝑇ℎ𝑢𝑠 𝑤ℎ𝑒𝑛 𝑦 = 2𝑥𝑒

−3𝑥

𝑑 𝑦 𝑑𝑦 , + 6 + 9𝑦 = 0 𝑑𝑥 𝑑𝑥2

1. 𝑰=∫

𝒅𝒙 𝟐𝒙𝟐 + 𝟐𝟎𝒙 + 𝟓𝟏

1 2𝑥 2 + 20𝑥 + 51 = 2(𝑥 2 + 10𝑥 + 25) + 1 = 2(𝑥 + 5)2 + 1 = 2 [(𝑥 + 5)2 + ] 2 𝐼=∫

𝑑𝑥

1 𝑑𝑥 = ∫ 2 1 2 [(𝑥 + 5)2 + ] 2 (𝑥 + 5)2 + ( 1 ) 2 √2 Let 𝑢 = 𝑥 + 5 ⇒ 𝑑𝑢 = 𝑑𝑥

1 𝐼= ∫ 2

𝑑𝑢 2

1 1 𝑢 √2 = . 1 𝑡𝑎𝑛−1 ( 1 ) + 𝑐 = 𝑡𝑎𝑛−1 (√2(𝑥 + 5)) + 𝑐 2 √2 2 √2

1 ) √2 1 = 𝑡𝑎𝑛−1 (√2(𝑥 + 5)) + 𝑐 √2

(𝑢)2 + (

2. 21 CALCULUS by Dr. Ibrahim

Differential Calculus by Dr. Syed Ibrahim  ⁄6

𝐼 = ∫ 4𝑠𝑖𝑛(4𝑡 + ⁄3) 𝑑𝑡 0

Let 𝑢 = 4𝑡 + ⁄3

 ⁄6

⇒ 𝑑𝑢 = 4𝑑𝑡

⁄  0  4  𝐼 = ∫ 𝑠𝑖𝑛𝑢 𝑑𝑢 = (−𝑐𝑜𝑠𝑢)0 6 = [𝑐𝑜𝑠 (4𝑡 + )] = 𝑐𝑜𝑠 ( ) − 𝑐𝑜𝑠 ( + ) 3 ⁄6 3 6 3 0

= 𝑐𝑜𝑠 ⁄3 − 𝑐𝑜𝑠 = 0.5 + 1 = 𝟏. 𝟓 3. 𝐼=∫

𝑥 √𝑥 + 1

𝐼=∫

𝑑𝑥

𝑢−1 √𝑢

𝑑𝑢 = ∫ (

Let

𝑢 =𝑥+1

𝑢

1

√𝑢

−

√𝑢

⇒ 𝑑𝑢 = 𝑑𝑥

) 𝑑𝑢 = ∫(𝑢

1⁄2

−𝑢

−1⁄2

𝑢3⁄2 𝑢1⁄2 − +𝑐 ) 𝑑𝑢 = 3 ⁄2 1⁄2 𝟑

2(𝑥 + 1)3⁄2 𝟐(√𝒙 + 𝟏) 𝐼= − 2(𝑥 + 1)1⁄2 + 𝑐 = − 𝟐√𝒙 + 𝟏 + 𝒄 3 𝟑 4. 5

𝐼 = ∫ (3𝑥 + 1) √3𝑥 2 + 2𝑥 + 5 𝑑𝑥 −1

Let 𝑢 = 3𝑥 2 + 2𝑥 + 5 5

5

⇒

𝑑𝑢 = (6𝑥 + 2)𝑑𝑥 5

⇒

𝑑𝑢 2

= (3𝑥 + 1)𝑑𝑥

𝑢3⁄2 2 5 1⁄2 𝐼 = ∫ √𝑢 𝑑𝑢 = ∫ 𝑢 𝑑𝑢 = [ ] = [(3𝑥 2 + 2𝑥 + 5)3⁄2 ]−1 3⁄ 3 2 −1 −1 −1 2 = [(90)3⁄2 − (6)3⁄2 ] = 𝟓𝟓𝟗. 𝟒 3

5. 22 CALCULUS by Dr. Ibrahim


𝐼 = ∫ 𝑡𝑎𝑛2 𝑥 𝑠𝑒𝑐 4 𝑥𝑑𝑥 sec2x = 1 + tan2x 𝐼 = ∫(𝑡𝑎𝑛2 𝑥) (𝑠𝑒𝑐 2 𝑥)(𝑠𝑒𝑐 2 𝑥)𝑑𝑥 = ∫(𝑡𝑎𝑛2 𝑥) (1 + 𝑡𝑎𝑛2 𝑥)(𝑠𝑒𝑐 2 𝑥)𝑑𝑥 Let 𝑢 = 𝑡𝑎𝑛𝑥 𝐼 = ∫𝑢

2 (1

+𝑢

2)

⇒ 𝑑𝑢 = 𝑠𝑒𝑐 2 𝑥 𝑑𝑥

𝑢3 𝑢5 𝑡𝑎𝑛3 𝑥 𝑡𝑎𝑛5 𝑥 𝑑𝑢 = + +𝑐 = + +𝑐 3 5 3 5

6. 𝐼 = ∫ 𝑥 √5 + 𝑥 2 𝑑𝑥 Let 𝑢 = 5 + 𝑥 2

⇒

𝑑𝑢 = 2𝑥𝑑𝑥

⇒ 𝑥𝑑𝑥 =

𝑑𝑢 2

𝑑𝑢 1 1 𝑢3⁄2 1 2 1 1⁄2 𝐼 = ∫ √𝑢 = ∫ 𝑢 𝑑𝑢 = + 𝑐 = . 𝑢3⁄2 + 𝑐 = 𝑢3⁄2 + 𝑐 2 2 2 (3⁄ ) 2 3 3 2 𝟏 = (𝟓 + 𝒙𝟐 )𝟑⁄𝟐 + 𝒄 𝟑 7. 1

𝐼 = ∫ 3𝑥 𝑒 (2𝑥

2 −1)

𝑑𝑥

0

Let 𝑢 = 2𝑥 2 − 1 1

⇒

𝑑𝑢 = 4𝑥 𝑑𝑥

⇒ 𝑑𝑥 =

1

𝑑𝑢 4𝑥

𝑑𝑢 3 3 3 3 1 2 𝐼 = ∫ 3𝑥 𝑒 𝑢 = ∫ 𝑒 𝑢 𝑑𝑢 = [𝑒 𝑢 ]10 = [𝑒 (2𝑥 −1) ]0 = [𝑒 1 − 𝑒 −1 ] 4𝑥 4 4 4 4 0

0

3 = (2.71 − 0.3678) = 1.7628 = 1.763 4 8.


Differential Calculus by Dr. Syed Ibrahim  ⁄3

𝐼 = ∫ 3𝑠𝑖𝑛2 3𝑥 𝑑𝑥 0

∵ 𝑠𝑖𝑛2  =

1 − 𝑐𝑜𝑠2 2

⇒

𝑠𝑖𝑛2 3𝑥 =

1 − 𝑐𝑜𝑠6𝑥 2

 ⁄3

3 3 𝑠𝑖𝑛6𝑥 ⁄3 3  𝑠𝑖𝑛2 𝑠𝑖𝑛0 𝐼 = ∫ (1 − 𝑐𝑜𝑠6𝑥) 𝑑𝑥 = [𝑥 − = [( − ] ) − (0 − )] 2 2 6 0 2 3 6 6 0

=

3   × = = 1.571 2 3 2

9. ∫ 2𝑠𝑖𝑛3𝑥 𝑠𝑖𝑛𝑥𝑑𝑥 1 sin 𝐴 sin 𝐵 = − [cos(𝐴 + 𝐵) − cos(𝐴 − 𝐵)] 2 𝑠𝑖𝑛4𝑥 𝑠𝑖𝑛2𝑥 ∫ 2𝑠𝑖𝑛3𝑥 𝑠𝑖𝑛𝑥𝑑𝑥 = − ∫(𝑐𝑜𝑠4𝑥 − 𝑐𝑜𝑠2𝑥) 𝑑𝑥 = − [ − ]+𝑐 4 2 𝑠𝑖𝑛2𝑥 𝑠𝑖𝑛4𝑥 =[ − ]+𝑐 2 4 10. 2

∫ 3𝑐𝑜𝑠8𝑡𝑠𝑖𝑛3𝑡 𝑑𝑡 1

1 cos 𝐴 sin 𝐵 = [sin(𝐴 + 𝐵) − sin(𝐴 − 𝐵)] 2



2

3 3 cos 11𝑡 cos 5𝑡 2 ∫ 3𝑐𝑜𝑠8𝑡𝑠𝑖𝑛3𝑡 𝑑𝑡 = ∫(sin 11𝑡 + sin 5𝑡) 𝑑𝑡 = (− + ) 2 2 11 5 1 1

1

2 2 3 cos 5𝑡 cos 11𝑡 2 3 3 = ( − ) = ( cos 5𝑡) − ( cos 11𝑡) 2 5 11 1 10 22 1 1 3 3 (cos 10 − cos 5) − (cos 22 − cos 11) = 10 22 3 3 (−0.8363 − 0.2836) − (−0.9999 − 0.0044) = 10 22 = −0.3359— 0.1369 = −𝟎. 𝟏𝟗𝟗𝟎
