Implicit Differentiation MCV4U: Calculus & Vectors Recap - PDF document

f u n d a m e n t a l r u l e s o f d e r i v a t i v e s f u n d a m e n t a l r u l e s o f d e r i v a t i v e s Implicit Differentiation MCV4U: Calculus & Vectors Recap Determine the derivative of y 3 + 4 x 2 y 5 − 5 x = 10. Remember that all terms involving y will have dy dx in their derivations, and that the central term uses the product rule. Higher-Order Derivatives y 3 + 4 x 2 y 5 − 5 x d = d � � dx 10 dx dx y 3 + 4 d J. Garvin d dx x 2 y 5 − 5 d dx x = 0 2 xy 5 + x 2 (5 y 4 ) dy � � 3 y 2 dy dx + 4 − 5 = 0 dx dx + 8 xy 5 + 20 x 2 y 4 dy 3 y 2 dy dx − 5 = 0 5 − 8 xy 5 dy dx = 3 y 2 + 20 x 2 y 4 J. Garvin — Higher-Order Derivatives Slide 1/17 Slide 2/17 f u n d a m e n t a l r u l e s o f d e r i v a t i v e s f u n d a m e n t a l r u l e s o f d e r i v a t i v e s Higher-Order Derivatives Higher-Order Derivatives Recall that the derivative of a function represents its rate of Example change. This is often called the first derivative . Determine the second derivative of f ( x ) = − 5 x 3 + 4 x 2 . What if we are interested in the rate of change of the Using the power rule, f ′ ( x ) = − 15 x 2 + 8 x . derivative itself – that is, how is the derivative changing with respect to the independent variable? Differentiating the first derivative, f ′′ ( x ) = − 30 x + 8. This concept is known as the second derivative . Example The Second Derivative Determine the second derivative of f ( x ) = √ x + 3. The second derivative is the derivative of the derivative 1 2 x − 1 function. In Lagrange notation, the it is denoted f ′′ ( x ). In 2 + 3, f ′ ( x ) = 1 2 . Using f ( x ) = x Leibniz notation, it is denoted d 2 y dx 2 . 4 x − 3 Differentiating this, f ′′ ( x ) = − 1 2 . To find the second derivative, differentiate the first derivative. J. Garvin — Higher-Order Derivatives J. Garvin — Higher-Order Derivatives Slide 3/17 Slide 4/17 f u n d a m e n t a l r u l e s o f d e r i v a t i v e s f u n d a m e n t a l r u l e s o f d e r i v a t i v e s Higher-Order Derivatives Higher-Order Derivatives Example Third, fourth or even higher-order derivatives can be calculated by further differentiation. √ 3 x 2 + 5, determine d 2 y If y = dx 2 . Higher-Order Derivatives � 1 3 x 2 + 5 2 to find dy � Use the chain rule on y = dx . In Lagrange notation, the n th derivative is denoted f ( n ) ( x ). 2 (3 x 2 + 5) − 1 In Leibniz notation, it is denoted d n y dy dx = 1 2 (6 x ) dx n . 3 x = The third derivative can be either f ′′′ ( x ) or f (3) ( x ) in 1 (3 x 2 + 5) 2 Lagrange notation, and d 3 y dx 3 in Leibniz notation. Use the quotient rule (or product and chain rules) to find d 2 y dx 2 . After the third derivative, prime notation becomes � 1 dx 2 = 3(3 x 2 + 5) 1 (3 x 2 + 5) − 1 2 − 3 x � 2 (6 x ) cumbersome for Lagrange notation, so numbers are used d 2 y 2 � 2 instead. � (3 x 2 + 5) 1 2 For instance, both d 9 y dx 9 and f (9) ( x ) are more readable than 15 f ′′′′′′′′′ ( x ). which simplifies to after quite a bit of algebra. 3 (3 x 2 + 5) 2 J. Garvin — Higher-Order Derivatives J. Garvin — Higher-Order Derivatives Slide 5/17 Slide 6/17

f u n d a m e n t a l r u l e s o f d e r i v a t i v e s f u n d a m e n t a l r u l e s o f d e r i v a t i v e s Higher-Order Derivatives Higher-Order Derivatives Example Example Determine the third derivative of f ( x ) = 6 x 5 − 3 x 2 + 5 x . If y = 2 x 99 , determine d 100 y dx 100 . Using the power rule, the first derivative is given by Recall that the power rule decrements the exponent by 1 f ′ ( x ) = 30 x 4 − 6 x + 5. each time it is applied. The second derivative is the derivative of f ′ ( x ), so For instance, the derivative of the quadratic function y = x 2 f ′′ ( x ) = 120 x 3 − 6. is a linear function dy dx = 2 x . The third derivative is the derivative of f ′′ ( x ), so The second derivative is d 2 y dx 2 = 2, a constant, and the third f ′′′ ( x ) = 360 x 2 . derivative is d 3 y dx 3 = 0. In general, a polynomial function of degree k will have a n th derivative of degree k − n . Since 99 − 100 < 0, then d 100 y dx 100 = 0. J. Garvin — Higher-Order Derivatives J. Garvin — Higher-Order Derivatives Slide 7/17 Slide 8/17 f u n d a m e n t a l r u l e s o f d e r i v a t i v e s f u n d a m e n t a l r u l e s o f d e r i v a t i v e s Displacement, Velocity and Acceleration Displacement, Velocity and Acceleration Velocity is the change of an object’s displacement with Example respect to time. Therefore, velocity is the derivative of the A ball is kicked off of a cliff and its height, h metres, after t seconds is given by h ( t ) = − 4 . 9 t 2 + 19 . 6 t + 24 . 5. With what displacement function. speed does the ball hit the ground? Velocity If s ( t ) represents an object’s displacement with respect to The ball hits the ground when h ( t ) = 0. time, then its velocity is given by v ( t ) = s ′ ( t ). In Leibniz − 4 . 9 t 2 + 19 . 6 t + 24 . 5 = 0 notation, v = ds dt . − 4 . 9( x + 1)( x − 5) = 0 An object is moving in a positive direction if its velocity is The ball hits the ground at 5 s. Find v (5) for its speed. positive, and in a negative direction if its velocity is negative. v ( t ) = − 9 . 8 t + 19 . 6 An object with a velocity of zero is at rest. v (5) = − 9 . 8(5) + 19 . 6 = − 29 . 4 Thus, the ball hits the round with a speed of 29 . 4 m/s. J. Garvin — Higher-Order Derivatives J. Garvin — Higher-Order Derivatives Slide 9/17 Slide 10/17 f u n d a m e n t a l r u l e s o f d e r i v a t i v e s f u n d a m e n t a l r u l e s o f d e r i v a t i v e s Displacement, Velocity and Acceleration Displacement, Velocity and Acceleration Acceleration is the change of an object’s velocity with Example respect to time. Therefore, acceleration is the derivative of The displacement, in metres, of a particle after t seconds is the velocity function. given by s ( t ) = − 2 t 3 + 8 t 2 − 10 t . • Determine the velocity of the particle at 2 seconds. Acceleration If s ( t ) represents an object’s displacement with respect to • When is the particle moving in a positive direction? In a time, and v ( t ) its velocity, then its acceleration is given by negative direction? At rest? a ( t ) = v ′ ( t ) or a ( t ) = d ′′ ( t ). In Leibniz notation, a = dv dt or • Determine the acceleration of the particle at 2 seconds. a = d 2 s dt 2 . • When is the particle moving with a constant velocity? An object with an acceleration of zero is moving at a An expression for the velocity of the particle is v ( t ) = s ′ ( t ) = − 6 t 2 + 16 t − 10. constant velocity. At t = 2, v (2) = − 6(2) 2 + 16(2) − 10 = − 2 m/s. Note that it is possible for an object to have a negative acceleration, but a positive velocity – it may be moving forward, but slowing down. J. Garvin — Higher-Order Derivatives J. Garvin — Higher-Order Derivatives Slide 11/17 Slide 12/17

f u n d a m e n t a l r u l e s o f d e r i v a t i v e s f u n d a m e n t a l r u l e s o f d e r i v a t i v e s Displacement, Velocity and Acceleration Displacement, Velocity and Acceleration Since v ( t ) is a quadratic function whose graph opens A graph of the displacement, s ( t ), and velocity, v ( t ), is downward, the particle will be moving in a positive direction below. for all values of t between the roots of the equation, and in a negative direction for all other values. v ( t ) = − 6 t 2 + 16 t − 10 = − ( t − 1)(6 t − 10) Since the roots are 1 and 5 3 , the particle is moving in a positive direction between 1 and 5 3 seconds. It is moving in a negative direction between 0 and 1 second, and for all times after 5 3 seconds. Therefore, the particle is moving in a positive direction before 3 seconds, and in a negative direction after 5 5 3 seconds. At 1 second and at 5 3 seconds the particle is briefly at rest. J. Garvin — Higher-Order Derivatives J. Garvin — Higher-Order Derivatives Slide 13/17 Slide 14/17 f u n d a m e n t a l r u l e s o f d e r i v a t i v e s f u n d a m e n t a l r u l e s o f d e r i v a t i v e s Displacement, Velocity and Acceleration Displacement, Velocity and Acceleration An expression for the acceleration of the particle is A graph of the velocity, v ( t ), and acceleration, a ( t ), is below. a ( t ) = v ′ ( t ) = − 12 t + 16. At t = 2 seconds, a (2) = − 12(2) + 16 = − 8 m/s 2 . An object moving at a constant velocity (possibly at rest) is not accelerating, so a ( t ) = 0. Therefore, the particle is moving at a constant velocity when − 12 t + 16 = 0, or at 4 3 seconds. J. Garvin — Higher-Order Derivatives J. Garvin — Higher-Order Derivatives Slide 15/17 Slide 16/17 f u n d a m e n t a l r u l e s o f d e r i v a t i v e s Questions? J. Garvin — Higher-Order Derivatives Slide 17/17