3.1. Derivatives of Power functions and Exponential functions ...

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Nowe we look at some examples: Example 1. Find the derivative of y = √ x − 2ex + x3 + 5. Solu- tion: First of all, rewrite the original function as follows to remove ...
3.1. Derivatives of Power functions and Exponential functions • Power function: y = xn , where n is a constant. √ • Note that Radical fuctions such as y = n x can be rewritten as 1 a Power function y = x n . • Exponential functions: y = ax , where a is a constant. When a = e, we have a special case of the Exponential function: y = ex . • Derivative of Power function: (xn )0 = nxn−1 . • (x5 )0 = 5x4 by using the Power function rule. • Derivative of Exponential function: (ex )0 = ex . • Sum rule for derivatives: (f (x) + g(x))0 = f 0 (x) + g 0 (x) • Difference rule for derivatives: (f (x) − g(x))0 = f 0 (x) − g 0 (x) • Constant multiple rule: (cf (x))0 = cf 0 (x), where c is a constant. Nowe we look at some examples: √ Example 1. Find the derivative of y = x − 2ex + x3 + 5. Solution: First of all, rewrite the original function as follows to remove the Radical: 1

y = x 2 − 2e2 + x3 + 5 Then we take derivative of y : 1

y 0 = (x 2 − 2ex + x3 + 5)0

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y 0 = (x 2 )0 − (2ex )0 + (x3 )0 + (5)0

1 1 y 0 = x− 2 − 2ex + 3x2 2 1

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3.2. Product rule and Quotient rule for derivatives • Product Rule: (f (x)g(x))0 = f 0 (x)g(x) + f (x)g 0 (x)

(x) 0 • Quotient Rule: ( fg(x) ) =

f 0 (x)g(x)−f (x)g 0 (x) (g(x))2

Now we look at some examples: Example 2. Find the derivative of f (x) = (2x2 + 3x + 1)ex . Solution: Since f (x) is a product of two functions, so we use the Product rule: (f (x))0 = ((22 + 3x + 1)ex )0

(f (x))0 = (2x2 + 3x + 1)0 ex + (2x2 + 3x + 1)(ex )0 (f (x))0 = (4x + 3)ex + (2x2 + 3x + 1)ex 2 2

Example 3. Find the derivative of f (t) = 2t3t+3t−5 3 +2 . Solution: Since f (t) is a quotient two functions, so we use the Quotient rule: f 0 (t) = (

f 0 (t) =

2t2 + 3t − 5 0 ) 3t3 + 2

(2t2 + 3t − 5)0 (3t3 + 2) − (2t2 + 3t − 5)(3t3 + 2)0 (3t3 + 2)2

f 0 (t) =

f 0 (t) =

(4t + 3)(3t3 + 2) − (2t2 + 3t − 5)(6t2 ) (3t3 + 2)2

(12t4 + 9t3 + 8t + 6) − (12t4 + 18t3 − 30t2 ) (3t3 + 2)2 f 0 (t) =

−9t3 + 30t2 + 8t + 6 (3t3 + 2)2 2

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3.3 Rates of change in the natural and social Sciences • In physics: Suppose s = f (t) be the function of position of a particle, then we have the following: – Velocity v(t) = f 0 is the derivative of the positiion function s = f (t). – Acceleration a(t) = v 0 (t) the derivative of velocity function.

• In Economics: Suppose C(q) = f (q) is the cost function for producing q units of product, then we have the following: – The derivative C 0 (q) is called the Marginal cost, which is the cost of producing the (q + 1)th unit of product. – Average cost of producing q units of product is given by C(q) q

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3.4 Derivatives of Trigonometric functions • • • • • • • • •

(sin x)0 = cos x (cos x)0 = − sin x (tan x)0 = sec2 x (cot x)0 = − csc2 x (sec x)0 = sec x tan x (csc x)0 = − csc x cot x limθ7→0 sinθ θ = 1 or limθ7→0 sinθ θ = 1 limθ7→0 cos θθ−1 = 0

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3.5 The chain rule • • • • • • • • • •

(f (x)n )0 = nf (x)n−1 f 0 (x) (ef (x) )0 = ef (x) f 0 (x) (af (x) )0 = (ln a)af (x) f 0 (x) (sin f (x))0 = cos f (x)f 0 (x) (cos f (x))0 = − sin f (x)f 0 (x) (tan f (x))0 = sec2 f (x)f 0 (x) (cot f (x))0 = − csc2 f (x)f 0 (x) (sec f (x))0 = sec f (x) tan f (x)f 0 (x) (csc f (x))0 = − csc f (x) cot f (x)f 0 (x) 0 dy If x = x(t), y = y(t), then dx = xy 0(t) (t)

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3.6 Implicit Differentiation • When you need to use implicit differentiation: If you have an equation about the function y and the variable x, and you want to know the derivative of y with respect to x, then you need to use implicit differentiation. • How to use implicit differentiation: The steps are as follows: – Take derivatives of both sides of the equation with respect to the variable x. You apply whatever rules necessary during this process. – Solve for y 0 . • When you are asked to find the derivative of inverse Trigonometric functions, you need to convert it into an equation first in terms of trigonometric functions, then use implicit differentiation.

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3.7 Derivatives of Logarithmic functions • • • •

The derivative of loga x is given as (loga x)0 = x ln1 a , (ln x)0 = x1 1 The corresponding chain rule: (ln f (x))0 = f (x) f 0 (x) Logarithmic differentation: The steps are as follows: – Step 1: Take ln of both sides and simply using properties of ln. – Step 2: Use implicit differentiation, then solve for y 0

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3. 8 Linear Approximation and Differentials • Linear Aproximation of f (x) is given by L(x) = f (a)+f 0 (a)(x− a). And we use L(x) = f (a)+f 0 (a)(x−a) to approximate f (x). • We should choose a in such a way that it is easy to compute f (a) and f 0 (a), and when x is very close to a, there is not a very big difference between L(x) and f (x) • Differential of y = f (x) is defined as dy = f 0 (x)dx, where f 0 (x) is the derivative of f (x) and dx = ∆x is the change of the variable. • We can use the differential dy to approximate ∆y which is the change of y. • The differential can be used to compute the error of the function caused by the error of the variable.