《C4.3 Differentiation and Integration 1.ppt》由会员分享，可在线阅读，更多相关《C4.3 Differentiation and Integration 1.ppt（36页珍藏版）》请在三一办公上搜索。
1、These icons indicate that teachers notes or useful web addresses are available in the Notes Page.,This icon indicates the slide contains activities created in Flash.These activities are not editable.,For more detailed instructions,see the Getting Started presentation.,Boardworks Ltd 2006,1 of 36,A2-
2、Level Maths:Core 4for OCR,C4.3 Differentiation and Integration 1,Differentiating trigonometric functions Implicit differentiationParametric differentiationExamination-style question,Contents,Boardworks Ltd 2006,2 of 36,Differentiating trigonometric functions,The derivative of sin x,The derivative of
3、 sin x,By plotting the gradient function of y=sin x,where x is measured in radians,we can deduce that,Functions of the form k sin f(x)can be differentiated using the chain rule.,Differentiate y=2 sin 3x with respect to x.,So if y=2 sin u whereu=3x,The derivative of sin f(x),Using the chain rule:,In
4、general using the chain rule,The derivative of cos x,The derivative of cos x,By plotting the gradient function of y=cos x,where x is measured in radians,we can deduce that,Find given that y=x2 cos x.,Letu=x2 andv=cos x,So,The derivative of cos x,Using the product rule:,Functions of the form k cos f(
5、x)can be differentiated using the chain rule.,Differentiate y=3 cos(x3 4)with respect to x.,So if y=3 cos u whereu=x3 4,The derivative of cos x,Using the chain rule:,In general using the chain rule,The derivative of tan x,We can differentiate y=tan x(where x is in radians)by writing it as,Then we ap
6、ply the quotient rule with u=sin x and v=cos x:,The derivative of sec x,We can differentiate y=sec x(where x is in radians)by writing it as,Then using the chain rule we get:,The derivative of cosec x,We can differentiate y=cosec x(where x is in radians)by writing it as,Then using the chain rule we g
7、et:,The derivative of cot x,We can differentiate y=cot x(where x is in radians)by writing it as,Then we apply the quotient rule with u=cos x and v=sin x:,Derivatives of trigonometric functions,In summary,if x is measured in radians,then,When learning these results,it is helpful to notice that all of
8、 the trigonometric functions starting with co have negative derivatives.,Differentiating trigonometric functions Implicit differentiationParametric differentiationExamination-style question,Contents,Boardworks Ltd 2006,15 of 36,Implicit differentiation,Implicit differentiation,When neither x nor y i
9、s given explicitly in terms of the other then the curve is said to be defined implicitly.,For example:,It is not always easy,or even possible,to rearrange an implicit function into an explicit form.,When the equation of a curve is given in the form y=f(x)then the variable y is said to be given expli
10、citly in terms of the variable x.,For example:,are explicit functions.,and,are implicit functions.,and,Implicit differentiation,Differentiating every term in the equation with respect to x gives:,For this reason we need to develop a technique to differentiate such functions in implicit form.,Differe
11、ntiate with respect to x.,where is taken as an operator meaning differentiate with respect to x.,The term in x and the constant can be differentiated directly to give:,Implicit differentiation,This would normally be done in a single step so:,To differentiate the term in y with respect to x we have t
12、o use the chain rule.,becomes,We can now divide through by 2y and rearrange to find:,Implicit differentiation,Differentiating term by term with respect to x gives:,Differentiate with respect to x.,Now rearrange to collect the terms in together:,3x2,Implicit differentiation,Differentiating term by te
13、rm with respect to x:,In some cases,we might also need to use the product rule.For example:,Differentiate with respect to x.,The first term is treated as a product to give:,Using,Implicit differentiation,Once we have differentiated a curve that has been defined implicitly,we can find the equation of
14、 the tangent or the normal to the curve at a given point.,Rearrange to collect the terms in together:,Implicit differentiation,Differentiating with respect to x gives:,Find the equation of the tangent to the curve x2+y2 xy=7 at the point(2,1).,Rearranging to collect the terms in together:,Implicit d
15、ifferentiation,At the point(2,1),x=2 and y=1:,The gradient of the tangent at(2,1)is therefore.,We can find the equation of the tangent using the coordinates:,Using y y1=m(x x1),Implicit differentiation,Differentiating with respect to x gives:,Find the equation of the normal to the curve 2x2(y+3)2=1
16、at the point(5,4).,Implicit differentiation,We need the gradient at the point(5,4):,The equation of the normal is given by:,Using y y1=m(x x1),The gradient of the normal at(5,4)is therefore.,Differentiating trigonometric functions Implicit differentiationParametric differentiationExamination-style q
17、uestion,Contents,Boardworks Ltd 2006,26 of 36,Parametric differentiation,Parametric differentiation,We can differentiate both of these equations with respect to the parameter t to give:,can be found using the chain rule:,Suppose we want to find for a curve that has been defined parametrically.,For e
18、xample:,and,Parametric differentiation,Differentiating each equation with respect to t gives:,In general,then,to differentiate a pair of parametric equations we can use the chain rule in the form:,Find,in terms of t,for the curve defined by the parametricequations x=cos 2t and y=sin t.,Parametric di
19、fferentiation,This can be simplified further using the double angle formula for sin 2t:,=cosec t,Parametric differentiation,Differentiating with respect to x gives,A curve is defined by the parametric equations x=(3t+2)2 and y=2t3+9.,Find in terms of t and hence find the coordinates of the points wh
20、ere the gradient of the curve is 1.,Parametric differentiation,The gradient of the curve is 1 when:,So the gradient of the curve is 1 when t=1 and when t=2.,When t=1:,x=(3+2)2,=1,and y=2(1)3+9,=7,When t=2:,x=(6+2)2,=16,and y=2(2)3+9,=7,The gradient of the curve is 1 at(1,7)and(16,7).,Differentiating
21、 trigonometric functions Implicit differentiationParametric differentiationExamination-style question,Contents,Boardworks Ltd 2006,32 of 36,Examination-style question,Examination-style question,Examination-style question,a),when t=1:,Using gives:,Examination-style question,b)The gradient of the norm
22、al to the curve at the point P is 1.,Also when t=1,If the gradient of the normal is 1 and it passes through the point(2,3)then its equation is given by:,x=2,y=3,c)Substituting x=2t2 and y=t2+2t into the equation for the normal gives:,Examination-style question,The normal therefore cuts the curve when t=1 and when t=.,When t=:,The coordinates of the point Q are therefore.,
