Differentiation
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.
the curve. finds the gradients of the tangents on Gradients of tangents and normals are related by The second derivative is. Which means differentiate
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) 7 8 7 i) - ) Differentiate with respect to a) b) c) - d) e) -
) Differentiate the following epressions with respect to a) + +7 b) - + c) + -7+ d) - + e) + - f) + -6 6) 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 =
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
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
. y = 7 + 0. (a) Find. () (Total marks). The curve C has equation y = + 8 +. 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
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
0. y A C R P O Q The diagram above shows part of the curve C with equation y = 6 + 8. The curve meets the y-ais at the point A and has a minimum at the point P. (a) Epress 6 + 8 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
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 = 6 + 8 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 = + 7 + (k) [k not required here] M A (, 0) = + + k k = 0 y = + 7 + 0 M A (d) = : = 8 + 7 = M A Gradient of normal = M y = ( ) + y = 0 M A [] d y. (a) 0 M A d (b) 7 C M A (, 0) []
. (a) =, y = 9 6 + + ( 9 6 + 7 = 0 is OK) B (b) + 8 (= 8 + 8) M A When =, = 9 + 8 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 8 + 8 = 7 M ( 8 + = 0) ( )( ) = 0 M = () or A = y = + 8 + 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)
6. (a) Evaluate gradient at = to get, Grad. of normal = B, M m Equation of normal: y ( ) (y = + 7) M A (b) ( ) = 9 6 + (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 8 8 6 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]
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]
Question EXTENSION QUESTIONS It is given that 8 y, 0. (a) Find the value of and the value of y when 0. [6]