www.mathsrevision.com Higher Outcome 2 Higher Unit 3 www.mathsrevision.com Further Differentiation Trig Functions Harder Type Questions Further Integration.

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Presentation transcript:

Higher Outcome 2 Higher Unit 3 Further Differentiation Trig Functions Harder Type Questions Further Integration Integration Exam Type Questions Integrating Trig Functions Differentiation The Chain Rule Harder Type Questions Differentiation Exam Type Questions

Higher Outcome 2 Trig Function Differentiation The Derivatives of sin x & cos x

Higher Outcome 2 Example Trig Function Differentiation

Higher Outcome 2 Example Trig Function Differentiation Simplify expression - where possible Restore the original form of expression

Higher Outcome 2 The Chain Rule for Differentiating To differentiate composite functions (such as functions with brackets in them) use: Example

Higher Outcome 2 The Chain Rule for Differentiating Trig Functions Trig functions can be differentiated using the same chain rule method. Worked Example:

Higher Outcome 2 The Chain Rule for Differentiating Example

Higher Outcome 2 Example The Chain Rule for Differentiating

Higher Outcome 2 Example The Chain Rule for Differentiating

Higher Outcome 2 The Chain Rule for Differentiating Trig Functions Example

Higher Outcome 2 Example The Chain Rule for Differentiating Trig Functions

Higher Outcome 2 Example Re-arrange: The slope of the tangent is given by the derivative of the equation. Use the chain rule: Where x = 3: The Chain Rule for Differentiating Functions

Higher Outcome 2 Is the required equation Remember y - b = m(x – a) The Chain Rule for Differentiating Functions

Higher Outcome 2 Example In a small factory the cost, C, in pounds of assembling x components in a month is given by: Calculate the minimum cost of production in any month, and the corresponding number of components that are required to be assembled. Re-arrange The Chain Rule for Differentiating Functions

Higher Outcome 2 Using chain rule The Chain Rule for Differentiating Functions

Higher Outcome 2 For x < 5 we have (+ve)(+ve)(-ve) = (-ve) Therefore x = 5 is a minimum Is x = 5 a minimum in the (complicated) graph? Is this a minimum? For x > 5 we have (+ve)(+ve)(+ve) = (+ve) The Chain Rule for Differentiating Functions For x = 5 we have (+ve)(+ve)(0) = 0 x = 5

Higher Outcome 2 The cost of production: Expensive components? Aeroplane parts maybe ? The Chain Rule for Differentiating Functions

Higher Outcome 2 Integrating Composite Functions By “reversing” the Chain Rule we see that: Worked Example

Higher Outcome 2 Integrating Composite Functions Example : Re-arrange Compare with standard form

Higher Outcome 2 Example Compare with standard form Integrating Composite Functions

Higher Outcome 2 Example Integrating Composite Functions

Higher Outcome 2 Example Integrating Composite Functions

Higher Outcome 2 Integrating Trig Functions Integration is opposite of differentiation Worked Example

Higher Outcome 2 From “reversing” the chain rule we can see that: Worked Example Integrating Trig Functions

Higher Outcome 2 Integrating Trig Functions Example Apply the standard form (twice). Break up into two easier integrals

Higher Outcome 2 Example Re-arrange Apply the standard form Integrating Trig Functions

Higher Outcome 2 Integrating Trig Functions (Area) Example A The diagram shows the graphs of y = -sin x and y = cos x a)Find the coordinates of A b)Hence find the shaded area C A S T 0o0o 180 o 270 o 90 o

Higher Outcome 2 Integrating Trig Functions (Area)

Higher Outcome 2 Integrating Trig Functions (Area)

Higher Outcome 2 Example Integrating Trig Functions Remember cos(x + y) =

Higher Outcome 2 Integrating Trig Functions

Higher Outcome 2 Example Use the “reverse” chain rule So we have: Giving: Integrating Functions

Differentiation Higher Mathematics Next

Calculus Revision Back Next Quit Differentiate

Calculus Revision Back Next Quit Differentiate Split up Straight line form Differentiate

Calculus Revision Back Next Quit Differentiate

Calculus Revision Back Next Quit Differentiate Straight line form Differentiate

Calculus Revision Back Next Quit Differentiate Multiply out Differentiate

Calculus Revision Back Next Quit Differentiate Straight line form Differentiate

Calculus Revision Back Next Quit Differentiate multiply out differentiate

Calculus Revision Back Next Quit Differentiate Straight line form Differentiate

Calculus Revision Back Next Quit Differentiate Straight line form multiply out Differentiate

Calculus Revision Back Next Quit Differentiate multiply out Simplify Differentiate Straight line form

Calculus Revision Back Next Quit Differentiate Chain rule Simplify

Calculus Revision Back Next Quit Differentiate Chain Rule Simplify

Calculus Revision Back Next Quit Differentiate Chain Rule

Calculus Revision Back Next Quit Differentiate Chain Rule Simplify

Calculus Revision Back Next Quit Differentiate Chain Rule Simplify

Calculus Revision Back Next Quit Differentiate Chain Rule Simplify Straight line form

Calculus Revision Back Next Quit Differentiate Chain Rule Simplify Straight line form

Calculus Revision Back Next Quit Differentiate Chain Rule Simplify Straight line form

Calculus Revision Back Next Quit Differentiate

Calculus Revision Back Next Quit Differentiate

Calculus Revision Back Next Quit Differentiate

Calculus Revision Back Next Quit Differentiate

Calculus Revision Back Next Quit Differentiate Chain Rule Simplify Straight line form

Calculus Revision Back Next Quit Differentiate Chain Rule Simplify

Calculus Revision Back Next Quit Differentiate Chain Rule Simplify Straight line form

Calculus Revision Back Next Quit Differentiate

Calculus Revision Back Next Quit Differentiate Chain Rule Simplify Straight line form

Calculus Revision Back Next Quit Differentiate Straight line form

Calculus Revision Back Next Quit Differentiate

Calculus Revision Back Next Quit Differentiate Chain Rule Simplify

Calculus Revision Back Next Quit Differentiate multiply out Differentiate

Calculus Revision Back Next Quit Differentiate Chain Rule Simplify

Calculus Revision Back Next Quit Differentiate Straight line form

Calculus Revision Back Next Quit Differentiate Straight line form Multiply out Differentiate

Calculus Revision Back Next Quit Differentiate Chain Rule Simplify

Calculus Revision Back Next Quit Differentiate

Calculus Revision Back Next Quit Differentiate Multiply out Differentiate

Calculus Revision Back Next Quit Differentiate Chain Rule Simplify

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Integration Higher Mathematics Next

Calculus Revision Back Next Quit Integrate Integrate term by term simplify

Calculus Revision Back Next Quit Find

Calculus Revision Back Next Quit Integrate Multiply out brackets Integrate term by term simplify

Calculus Revision Back Next Quit Find

Calculus Revision Back Next Quit Integrate Standard Integral (from Chain Rule)

Calculus Revision Back Next Quit Find p, given

Calculus Revision Back Next Quit Evaluate Straight line form

Calculus Revision Back Next Quit Find Use standard Integral (from chain rule)

Calculus Revision Back Next Quit Find Integrate term by term

Calculus Revision Back Next Quit Integrate Straight line form

Calculus Revision Back Next Quit Integrate Straight line form

Calculus Revision Back Next Quit Integrate Straight line form

Calculus Revision Back Next Quit Integrate Split into separate fractions

Calculus Revision Back Next Quit Find Use standard Integral (from chain rule)

Calculus Revision Back Next Quit Find

Calculus Revision Back Next Quit Find

Calculus Revision Back Next Quit Integrate Straight line form

Calculus Revision Back Next Quit Given the acceleration a is: If it starts at rest, find an expression for the velocity v where Starts at rest, so v = 0, when t = 0

Calculus Revision Back Next Quit A curve for which passes through the point Find y in terms of x. Use the point

Calculus Revision Back Next Quit Integrate Split into separate fractions Multiply out brackets

Calculus Revision Back Next Quit If passes through the point express y in terms of x. Use the point

Calculus Revision Back Next Quit Integrate Straight line form

Calculus Revision Back Next Quit The graph of passes through the point (1, 2). express y in terms of x. If simplify Use the point Evaluate c

Calculus Revision Back Next Quit Integrate Straight line form

Calculus Revision Back Next Quit A curve for which passes through the point (–1, 2). Express y in terms of x. Use the point

Calculus Revision Back Next Quit Evaluate Cannot use standard integral So multiply out

Calculus Revision Back Next Quit Evaluate Straight line form

Calculus Revision Back Next Quit Evaluate Use standard Integral (from chain rule)

Calculus Revision Back Next Quit The curve passes through the point Find f(x ) use the given point

Calculus Revision Back Next Quit Integrate Integrate term by term

Calculus Revision Back Next Quit Integrate Integrate term by term

Calculus Revision Back Next Quit Evaluate

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