Calculus 105 AAG Lecture Notes - Function Derivatives PDF

Agronomy Dept. Calculus 105 AAG Lec 1. Part II Introduction to Function Derivatives by: Dr. Ibrahim yacoub  Contents Definition of the derivative of a function How to compute the derivative of a function based upon the definition ?...

Agronomy Dept. Calculus 105 AAG Lec 1. Part II Introduction to Function Derivatives by: Dr. Ibrahim yacoub  Contents Definition of the derivative of a function How to compute the derivative of a function based upon the definition ? Some proprieties of differentiation methods Importance of derivatives ? Examples  Plant Hight cm Y=f(x) 15 10 f(x+∆x) ∆y= y2 – y1 5 f(x) x 1 2 Age week ∆X= x –x 2 1  Plant Hight cm 20 f(x+∆x) ∆y= y2 – y1 5 f(x) 1 2 Age week ∆X= x –x 2 1 What is the Derivative of a Function ? The derivative of a function f(x) represents its rate of change & is denoted by either f'(x) or df/dx. About the Notation The derivative of a function can be denoted by both f'(x) and dy/dx. The mathematical giant Newton used f'(x) to denote the derivative of a function. Leibniz, another mathematical hero, used dy/dx. So dy/dx is a single term, not to be confused with a fraction. It is read as the derivative of a function y with respect to x , and also indicates that x is the independent variable. Slope of a Curve: A Geometrical Approach to the derivatives We have seen that the slope of a line can be written: Where Δy =y2 – y1 is a finite change in y and Δx =x2 – x1 is a finite change in x. The slope of a line can be found by taking any two points on a line as it remains the same at every point on the line. (hypotenuse) (opposite) θ (adjacent) B ( x +  x , f ( x +  x )) x x  x x The slope of the tangent to the curve y = f(x) is given by: The slope of the tangent at P is obtained if Q→P i.e if Δx→0 We say that in the limit as Δx approaches zero, the slope of the tangent to the curve at the point (x , y) is given by The slope of the tangent to a curve at a point is a limit. dy The whole expression for the limit is written , the dx derivative of y with respect to x.  Differentiation The method of finding the derivative of a function is called differentiation. In this section, we’ll see how the definition of the derivative can be used to find the derivative of different functions.  Differentiation Examples Example 1: Example 2: Example 3: m(x) = 2x+5 g(x) = x2 h(x) = x3 Ex. ( 5 ) Find Dx y if y = x2 at any point. Sol The value of this function at any point P is equal to Np = x2 Its value at P` is: Mp`=(x+x)2  y = (x +  x ) − x = 2 x  x + ( x ) 2 2 2 Dividing by x, then y = 2x + x x D x y = lim (2x + x ) x → 0 D x y = 2x Review for some principal rules of differentiation 1) Derivative of constant function: Ex. If y = 5 then f ( x ) = 0 Ex. 2 dy If y = then =0 3 dx 2) The derivative of a constant times a function: d dv (c.v ) = c dy dx Ex. If f ( x ) = 10 x 3 , then d ( dx ) d 3 10 x = 10 x 3 dx = 30 x 2 3) The derivative of a power function: b −1 If f ( x ) = x b , then f ( x ) = b x d 5 Ex. t = 5t 4 dt 4) The derivative of a polynomial function: d du dv (u  v ) =  dx dx dx If y = (x2 + c) + (ax4 + b) Then y`= 2x + 4 ax3 5) The derivative of a product: d (u.v ) = u + v dv du dx dx dx Ex. If y = x3 sin x Then y`= x3 cos x + 3 x2 sin x 6) The derivative of Quotient: du dv v −u d u dx dx   = dx  v  v 2 bx 5 + c Ex. If y= 2 x +a let bx 5 + c = u and x 2 + a = v  Sol ( x +a. 2 d bx 5 ) + ( c − bx + c 5 d x 2 ) ( +a ) ( ) dy dx dx = dx x +a 2 2 ( ) = (x 2 )( ) ( + a 5bx − bx + c (2x ) 4 5 ) (x + a) 2 2 3bx 6 + 5abx 4 − 2cx = (x 2 +a ) 2 7) Differentiation of a function of a function dy dy dv =. dx dv dx Ex. : If : y = v2 − 3 v + 2 and v = 4 x2 + 1 Find dy dx Sol. dy dv =2v − 3 , = 8x dv dx dy = ( 2v − 3 ). ( 8 x ) = 8 x ( 8 x 2 − 1 ) dx Ex. Use the power rule to find Dxy where; ( ) 3 y= x +a 2 2 2 v = x2 3+ a2 Sol. y = (v) 2 dy 3 1 dv = v 2 , = 2x dv 2 dx dy dy dv 3 1 =. = ( v 2 ) (2 x ) dx dv dx 2 3 1 1 = (x + a ) 2 2 2 ( 2 x) = 3 x ( x + a ) 2 2 2 2  Differentiation of parametric equations x = 2t + 3 , y = t2 − 1 dy To find: dx dy dy dx =. dt dx dt If ( d x / d t  0 ) dy dy dx Then = / dx dt dt Sol. dx dy =2 , = 2t dt dt dy dy dx x−3  = / = 2t / 2 = t = dx dt dt 2  Differentiation of inverse function y If y = 3 x The inverse function is: =x 3 dy y = 3x  =3 dx y dx 1 x=  = 3 dy 3 dy dx dy dx =1/ =3 Or. =1 dx dy dx dy Importance of derivatives ! 2y2x= 22x 3 − 2 x + 2 y 2= x x Quiz 1 choose the correct answers