compound angle identity for sin (A + we find that B) By manipulation we obtain Since 0, and we conclude that → Therefore the gradient of the chord gradient of the curve and we conclude that . → A similar argument with = cos as a starting point leads to: y x and therefore finding the derivative to be x. –sin The alternative notations 0 rather than 0 are acceptable. h → → OPPORTUNITIES FOR REASONING/PROBLEM SOLVING Ask the students to experiment with a graph-drawing package to verify that the gradient functions of sin x and cos match the result found using first principles. Students must understand that the differentiation of x sin and cos can only be used when is in radians and that they must use radians whether stated in the x x x question or not. NOTES The rest of this unit covers differentiation of more complicated functions in which the derivatives of sin x and cos are building blocks. x Teaching time 8b. Differentiating exponentials and logarithms (7.2) 3 hours OBJECTIVES By the end of the sub-unit, students should: be able to differentiate functions involving ex, ln and related functions such as 6e4x and 5 ln 3x x and sketch the graphs of these functions; be able to differentiate to find equations of tangents and normals to the TEACHING POINTS It is vital that students understand the functions ex and ln and do not just learn how to differentiate them. x Use a graphing tool to show that the gradient of a special curve = has a gradient which is exactly y ax ax. In other words its rate of growth is exactly the same as its value at that point. This models biological growth in nature (and decay if we consider The curve sits between 2x and 3x and has a value of a–x) 2.718… We call this exponential e. Therefore if = ex, = ex. y Explain that if = 2ex then = 2ex. y The students could verify this on the graphs below as Fig. 1 is effectively a stretch parallel to the y-axis. Fig. 2 shows that the graph of = e2x is twice as steep as ex, hence if = e2x then = 2e2x. y y These results will be deduced more formally in Unit 8c.
compound angle identity for sin (A + B) we find that By manipulation we obtain Since → 0, and we conclude that Therefore the gradient of the chord → gradient of the curve and we conclude that . A similar argument with y = cos x as a starting point leads to: and therefore finding the derivative to be –sin x. The alternative notations h → 0 rather than → 0 are acceptable. OPPORTUNITIES FOR REASONING/PROBLEM SOLVING Ask the students to experiment with a graph-drawing package to verify that the gradient functions of sin x and cos x match the result found using first principles. Students must understand that the differentiation of sin x and cos x can only be used when x is in radians and that they must use radians whether stated in the question or not. NOTES The rest of this unit covers differentiation of more complicated functions in which the derivatives of sin x and cos x are building blocks. 8b. Differentiating exponentials and logarithms (7.2) Teaching time 3 hours OBJECTIVES By the end of the sub-unit, students should: be able to differentiate functions involving ex, ln x and related functions such as 6e4x and 5 ln 3x and sketch the graphs of these functions; be able to differentiate to find equations of tangents and normals to the TEACHING POINTS It is vital that students understand the functions ex and ln x and do not just learn how to differentiate them. Use a graphing tool to show that the gradient of a special curve y = ax has a gradient which is exactly ax. In other words its rate of growth is exactly the same as its value at that point. This models biological growth in nature (and decay if we consider a–x) The curve sits between 2x and 3x and has a value of 2.718… We call this exponential e. Therefore if y = ex, = ex. Explain that if y = 2ex then = 2ex. The students could verify this on the graphs below as Fig. 1 is effectively a stretch parallel to the y-axis. Fig. 2 shows that the graph of y = e2x is twice as steep as ex, hence if y = e2x then = 2e2x. These results will be deduced more formally in Unit 8c.