Formulae to Learn. The Rules for Differentiation are. The instructions are to either: Find ( or ), or

Transcription

1 Differentiation

2 Formulae to Learn The Rules for Differentiation are The instructions are to either: Find ( or ), or Differentiate, or Find the derived function, or Find the derivative.

3 the curve. finds the gradients of the tangents on Gradients of tangents and normals are related by The second derivative is. Which means differentiate

4 STARTER QUESTIONS ) Find the derived function of the following a) b) c) 6 d) 7 e) 8 f) 6 8 g) h) i) 7 j) - ) Differentiate a) b) 7 c) d) e)- f) 8 g) h) - i) 7 7 ) Differentiate with respect to a) b) 7 c) 7 d) e) -7 f) - g) h) i) - ) Differentiate with respect to a) b) c) - d) e) -

5 ) Differentiate the following epressions with respect to a) + +7 b) - + c) d) - + e) + - f) ) Find the gradient of the curve whose equation is a) y= at the point (, ) b) y= -+ at the point (, ) c) y= + - when =- d) y= 7- - when =

6 STARTER QUESTIONS ) Simplify and differentiate the following with respect to a) y= ( +8) b) y= ( +) c) y= ( - +) d) y= e) y= ( + )( ) f) y= ( )( ) g) y= h) y= i) y= j) y= k) y= l) y= m) y= n) y= ) For each of the following epressions find d y and a) y= + b) y= + - c) y= d) y= 6

7 Past Paper Questions (some of the following questions are part of longer questions). (i) Given that y = + 7 +, find (a), d y (b). () () (Total marks). Given that y = 6, 0, (a) find, () (Total marks). For the curve C with equation y = f(), = + 7. d y (a) Find. d y (b) Show that for all values of. () () Given that the point P(, ) lies on C, (c) find y in terms of, () (d) find an equation for the normal to C at P in the form a + by + c = 0, where a, b and c are integers. () (Total marks) 7

8 . y = (a) Find. () (Total marks). The curve C has equation y = The point P has coordinates (, 0). (a) Show that P lies on C. (b) Find the equation of the tangent to C at P, giving your answer in the form y = m + c, where m and c are constants. () () Another point Q also lies on C. The tangent to C at Q is parallel to the tangent to C at P. (c) Find the coordinates of Q. () (Total marks) 6. The gradient of the curve C is given by = ( ). The point P(, ) lies on C. (a) Find an equation of the normal to C at P. () (b) Find an equation for the curve C in the form y = f(). () (c) Using = ( ), show that there is no point on C at which the tangent is parallel to the line y =. () (Total marks) 8

9 7. The curve C with equation y = f() is such that = +, > 0. (a) Show that, when = 8, the eact value of is 9. () The curve C passes through the point (, 0). (b) Using integration, find f(). (6) (Total 9 marks) 8. The curve C has equation y = +, 0. The point P on C has -coordinate. (a) Show that the value of at P is. (b) Find an equation of the tangent to C at P. () () 9. This tangent meets the -ais at the point (k, 0). (c) Find the value of k. () (Total 0 marks) C R O A The curve C, with equation y = ( ), intersects the -ais at the origin O and at the point A, as shown in the diagram above. At the point P on C the gradient of the tangent is. (a) Find the coordinates of P. () (Total marks) 9

10 0. y A C R P O Q The diagram above shows part of the curve C with equation y = The curve meets the y-ais at the point A and has a minimum at the point P. (a) Epress in the form ( a) + b, where a and b are integers. () (b) Find the coordinates of P. (c) Find an equation of the tangent to C at A. The tangent to C at A meets the -ais at the point Q. () () (d) Verify that PQ is parallel to the y-ais. () (Total 0 marks) 0

11 Past Paper Solutions. (i) (a) + 7 M A A (i) (b) 0 Bft (ii) C A: + C, A: +, A: + M A A A [8]. (a) d y = M A M is for n n in at least one term, 6 or is sufficient. A is fully correct answer. Ignore subsequent working. [] d y. (a) = + M A (b) Since d y is always positive, for all. B (c) y = (k) [k not required here] M A (, 0) = + + k k = 0 y = M A (d) = : = = M A Gradient of normal = M y = ( ) + y = 0 M A [] d y. (a) 0 M A d (b) 7 C M A (, 0) []

12 . (a) =, y = ( = 0 is OK) B (b) + 8 (= 8 + 8) M A When =, = m = 7 M Equation of tangent: y 0 = 7( ) M y = 7 + A c.a.o. st M some correct differentiation ( n n for one term) st A correct unsimplified (all terms) nd M substituting p (= ) in their clear evidence rd M using their m to find tangent at p. (c) = m gives = 7 M ( 8 + = 0) ( )( ) = 0 M = () or A = y = M 6 y = or A [] st M forming a correct equation their = gradient of their tangent nd M for solving a quadratic based on their leading to =... The quadratic could be simply = 0. rd M for using their value (obtained from their quadratic) in y to obtain y coordinate. Must have one of the other two M marks to score this. MR For misreading (0, ) for (, 0) award B0 and then MA as in scheme. Then allow all M marks but no A ft. (Ma 7)

13 6. (a) Evaluate gradient at = to get, Grad. of normal = B, M m Equation of normal: y ( ) (y = + 7) M A (b) ( ) = (May be seen elsewhere) B 9 6 Integrate: ( C) M Aft Substitute (, ) to find c =, c = (y = + + ) M, Acso (c) Gradient of given line is B Gradient of (tangent to) C is 0 (allow >0), so can never equal. B [] 7. (a) 8 = seen or used somewhere (possibly implied). B 8 or Direct statement, e.g. (no indication of method) is M0. M At = 8, =8 + = 6 + = 9 (*) A 8 (b) Integrating: ( C) (C not required) M A A At (, 0), C 0 (C required) M (f() =), A, A 6 [9] 8. (a) M = 8, M A, A When =, = (*) A cso (b) At P, y = 8 B Equation of tangent: y 8 = ( ) (y = + ) (or equiv.) M Aft (c) Where y = 0, = (= k) (or eact equiv.) M A [0]

14 9. (a) y = = M A =, = M =, y = A [] 0. (a) ( ), +9 isw. a = and b = 9 may just be written down with no method shown. B, M A (b) P is (, 9) B (c) A = (0, 8) B = 6, at A m = 6 M A Equation of tangent is y 8 = 6 (in any form) Aft (d) Showing that line meets ais directly below P, i.e. at =. Acso [9]

15 Question EXTENSION QUESTIONS It is given that 8 y, 0. (a) Find the value of and the value of y when 0. [6]