DHS Differentiation & Its Applications (9758) (Revision Solutions)

Y5 Topical Revision Package 1 4. Differentiation and its Applications (solutions) 1 4. Differentiation and its Applications (solutions) (I) Tangents and Normals (Normal/Implicit Differentiation) 1 (a) (b) ( ) ( )( ) ( )( ) 1 1 212 221 12 Let sin 3 3sin (3 ) d 1 9 sin (3 ) 319 d sin 3 sin 3 1 9 xy x x x y x xxx x x x x − − − −− = −  −−−= = − ( )( ) cos 2 3 3 Taking ln on both sides, cos 2 ln ln 3ln Differentiate w.r.t. , xyx xy x x x = = = ( )( ) ( ) ( ) ( ) ( ) cos2 3 3 3 3 1d 32sin2 ln cos2 ........... (1) d At , 1d 3Subst into (1): 2sin2 ln cos2 d yxy x yx x xy y y x −+= = = ⇒= −+= ππ ππ ππ π ππ 2d 3d y x π∴= 2 (i) (ii) (iii) ( )de e e e 1d x x xxyyx x xx − − −−= ⇒ = −= − Graph is decreasing: d 0d y x < ( )e 1 0 1x xx− −< ⇒ > [Also accept 1x≥ ] ( ) ( ) ( ) 2 2 d e 1e 1 e 2 d xx xy xx x −− −= −− − = − Graph is concave downwards: 2 2 d 0 d y x < ( )e 2 0 2x xx− −< ⇒ < Therefore, for graph to be decreasing and concave downwards: 12 x<< . [Also accept 12 x≤< , 12 x<≤ , or 12 x≤≤ ] Consider gradient at ( ),ab = ( )e1a a− − : ( ) ( ) e10 e e1 a aa hb aa ha a a − −− − = −− −= − − Since e aba −= ( ) 2e e1 eaa aha a a a−− −= − −= Note There is no need to simplify d d y x expression. We need the corresponding y-coordinate to evaluate the gradient to the curve at x π= . 1x> 2x< Alternative Tangent at ( ),ab : ( )( )e2ayb a xa −−= − − Subst ( )0,Rh into tangent and e aba −= : ( ) 2e e1 eaa aha a a a−− −= − −=

Y5 Topical Revision Package 1 4. Differentiation and its Applications (solutions) 2 ( )2d 2e e e 2d a aah aa a aa − −−=−= − At max/min point: d 0d h a = ( )e2 0 2 or 0 aaa aa − −= = = a 2 – 2 2+ d d h a + 0 – Slope a 0 – 0 0+ d d h a – 0 + Slope Alternative : Second derivative Test ( ) 2 2 2 ed 24 d 0 ah aa a − = −+ < when 2a= Greatest possible h 24e−= at 2.a= 3. (i) (ii) f( )yx= strictly increasing ⇒ f '( )x > 0 f( )yx= concave downwards ⇒ f ''( ) 0x < [so the gradient function f '( )x decreases] We observe from f '( )yx= graph that 1x> (ans) For stationary points on graph of f( )yx= , Need f '( ) 0x = ⇒ 3, 0x=− x (−3)− −3 (−3)+ f '( )x −ve 0 −ve tang Point of inflexion at x = −3 x 0− 0 0+ f '( )x −ve 0 +ve tang Minimum point at x = 0. [We observe the signs from f '( )yx= graph] (iii) Alternative : From the graph, greatest h occurs when d d y x is most negative (i.e. where d d y x is min, in this case). Thus need 2 2 d d 02y x x=⇒= Greatest possible 22 2 42e eh −= = x y 1 −3 −1.5 x =1

Y5 Topical Revision Package 1 4. Differentiation and its Applications (solutions) 3 4 (i) (ii) (iii) Differentiating 111 xy