MATH141: Mathematics 1C Part 1

Overview of Lectures on differentiation

In this section I will try to summarise the main ideas of each lecture.

Week 5, lecture 2 (Functions).

Exam questions on functions (from the differentiation part of the course, not the fundamentals part of the course).

There are video resources on "functions" on the Mathematics Skills Review (goto the e-learning page). (Some of the functions questions cover material from the fundamentals part of the course).

1. What is a function?
2. What are the domain and range of a function?
3. How do you find the domain and range of a function f(x) from the graph y=f(x) ?
4. Given the graph of the function y=f(x) explain how to draw the graphs of the functions: y=f(x-a), y=f(x)+a and y=1/f(x).
5. Given the domain of functions f(x) and g(x) what are the domains of the following functions:
   (i) f(x)+g(x),
   (ii) (fg)(x),
   (iii) (f/g)(x)
   (iv) αf(x), for some real number α

Week 6, lecture 1 (Limits).

• I've written a worksheet on limits (solutions).
• Exam questions on limits.
• There are video resources on "limits" on the Mathematics Skills Review (goto the e-learning page). (Nine questions the last time I looked).

1. The key idea about limits is that the limit of a function as x approaches a point does not depend on the value of the function at the point.
2. Consider a rational function r(x) = p(x)/q(x). If
   1. lim_x->ar(x) = 0/0 then we should factorise the expression and re-calculate the limit. (This is one-reason why you need to be able to factorise).
   2. lim_x->ar(x) = ∞/∞ then we should divide the numerator and denominator of the expression by the highest power of x in r(x). Then re-calculate the limit.
   3. lim_x->ar(x) = 1/x then the limit is zero.

Week 6, lecture 2 (Elementary differentiation).

• Worksheet on elementary differentiation (solutions).
• Exam questions on elementary differentiation.
• There are video resources on differentiation on the Mathematics Skills Review (goto the e-learning page). (The video resources in this section also cover more advanced differentiation topics).

1. Differentiation of simple functions such as c (a constant), xⁿ, sin(x), cos(x), tan(x), e^x and ln(x).
2. The product rule
   (fg)'(x) = g(x)f'(x) +f(x)g'(x).
3. The quotient rule
   (f/g)'(x) = (g(x)f'(x) -f(x)g'(x))/[g(x)]²
4. The chain rule. If y = f(g(x)) then
   y' = f'(g(x)).g'(x)
   (There's another way to write the chain rule which is slightly more useful, but I haven't had time to figure out how to typeset it in HTML).

Week 7, lecture 1 (Hyperbolic functions; one-to-one functions and inverse functions).

Exam questions on hyperbolic functions and one-to-one functions and inverse functions.

There are video resources on hyperbolic functions on the Mathematics Skills Review (goto the e-learning page). These questions appear in the differentiation section and in the "hyperbolic functions" section. (There were six questions in the latter the last time I looked). There are video resources on one-to-one functions and inverse functions on the Mathematics Skills Review (goto the e-learning page). These resources are in the "functions" section.

1. Definition of sinh(x), cosh(x) and tanh(x).
2. Graphs of hyperbolic functions, including domain and range.
3. cosh²(x) -sinh²(x) = 1 and related identities.
4. If you are asked to prove an hyperbolic identity... don't panic! The following strategies will be used in most questions:
   • Substitute for hyperbolic functions, i.e.
     cosh(x) = (1/2)*(e^x +e^-x)
     sinh(x) = (1/2)*(e^x -e^-x)
   • Use the identity that
     cosh²(x) -sinh²(x) = 1
5. Derivatives of hyperbolic functions.
6. Definition of one-to-one function.
7. The functions f(x) and g(x) are inverses of each other if
   f(g(x)) = x and g(f(x))=x.

Week 7, lecture 2 (one-to-one functions and inverse functions continued)

1. Under what circumstances does the function y = f(x) have an inverse?
2. How do you find the domain and range of an inverse function?

   The answer to this question uses a big idea. Inverse functions are difficult to grasp but functions are easy to grasp. So the turn the question around: make it a question about the function.

3. Given a function f(x), what is the procedure to find its inverse?

Week 8, lecture 1 (inverse trigonometry functions)

1. Why do we need to restrict the domain of trig functions?
2. What are the domain and range of the functions: arcsin(x), arccos(x) and arctan(x).
3. Don't forget the big idea. Inverse functions are difficult to grasp but functions are easy to grasp. So consider turning a question about an inverse trig function around: make it a question about a trig function.

Week 8, lecture 2 (inverse hyperbolic functions, the derivative of an inverse function, Logarithmic differentiation, implicit differentiation)

• I've written a worksheet on implicit differentiation (solutions).

(inverse hyperbolic functions)

1. What are the domain and range of the functions: arcsinh(x), arccosh(x) and arctanh(x).
2. There are formulae for arcsinh(x), arccosh(x) and arctanh(x) in terms of . Put these onto your formulae sheet now!

(the derivative of an inverse function)

3. How to differentiate an inverse function? Turn the question around by applying the function to both sides and then differentiating the function. This is one of the big ideas.

(Logarithmic differentiation)

4. To differentiate the function y = f(x)^g(x) we apply the following procedure.
   y = f(x)^g(x)
   y = exp[ ln(f(x)^g(x))]
   y = exp[ g(x)ln(f(x))]
   y'= f(x)^g(x) × [g(x)ln(f(x))]'

(implicit differentiation) The Chain Rule revisited!

5. How to find y' if y = f(g(x))?
   Define u=g(x) so that y=f(u).
   Then u_x = g'(x) and y_u = f'(u).
   y_x = u_x × y_u
   y_x = g'(x)× f'(g(x)).

Week 9, lecture 1 (implicit differentiation continued; parametric equations and curves)

• I've written a worksheet on parametric differentiation (solutions).

(implicit differentiation)

1. Suppose we want to find dy/dx given the equation x² -2y³ +4y =2. Suppose that we are given a value for x. Then we can calculate the corresponding value for y. So y is a function of x.
2. The key idea is that we can't differentiate a function of y with respect to x, but we can use the chain rule to differentiate a function of y with respect to y.
   d/dx(g(y)) = d/dy(g(y))×dy/dx.
3. Given an equation involving x and y we can find dy/dx as follows.
   1. Differentiate all terms of the equation with respect to x. Use the chain rule to differentiate all functions involving y.
   2. Collect all terms involving dy/dx on the left hand size of the equation and move all other terms to the right side.
   3. Take dy/dx out as a factor on the left hand side of the equation.
   4. Solve for dy/dx by dividing both sides of the equation by the left hand factor that multiples dy/dx.
4. To find d²y/dx² we need to.
   1. Remember that
      d²y/dx² = d/dy (dy/dx).
      Now dy/dx is a function of both y and x. We will need to use implicit differentiation when we differentiate dy/dx.
   2. Remember that when we differentiate a function of y, say g(y), we obtain d/dy(g(y))×dy/dx, that we know what dy/dx is and should substitute for it.
5. Show, using implicit differentiation, that x(x²+3y²) =c is the solution to the differential equation
   x² +y² +2xydy/dx = 0.
   To do a question like this:
   1. Use implicit differentiation to find dy/dx.
   2. Substitute your expression for dy/dx into the differential equation and simplify. If LHS=RHS then you have shown that the given expression is indeed a solution to the differential equation.

Parametric equations and curves. Sometimes we are given x and y as a function of a parameter t, i.e.
x = f(t)
y = g(t).
For a parametric system you need to be able to:

• sketch the parametric curve in the y-x plane;
• to find dy/dx
• to find d²y/dx².

6. To sketch the parametric curve in the y-x plane draw up a table containing t, y and x. For each value of t in the table calculate the corresponding values of x and y using the parametric equations
   x = f(t)
   y = g(t).
   You may also be able to eliminate t from the equations and derive an explicit relationship between y and x.

Week 9, lecture 2 (parametric equations and curves continued)

1. To find dy/dx we:
   1. calculate dy/dt and dx/dt;
   2. use the chain rule to find dr/dx from
      dy/dx = dy/dt× dt/dx
      dy/dx = dy/dt/(dx/dt).
2. To find d²y/dx² we need to remember that
   d²y/dx² = d/dx(dy/dx).
   Now dy/dx is a function of the parameter t. We need to use the chain rule to write:
   d²y/dx² = d/dt(dy/dx)×dt/dx
   d²y/dx² = d/dt(dy/dx)/(dx/dt).