C1 Differentiation 1g55k

C1 Differentiation Note: Bold means an important term you should understand, Red means a worked example, Red, bold and underlined means an answer, Yellow highlighting means an important exam tip. All the worked examples come from actual C1 papers. A few you should : These are some "buzzwords" related to differentiation so you should them well!  Function: A relationship between two variables (e.g. x and y) expressed in an equation or on a graph, e.g. y = 2x, y = sin x, etc. (There's a more accurate definition in C3 but that's all you'll need to know for C1.)  Gradient: The slope of a line, the change in Y divided by the change in X: Recall that the equation for a straight line is y = mx + c, where m is the gradient of the line.  Tangent: A line that just touches a curve at one point. The gradient of the tangent is the gradient of the curve at that point. (A secant by the way is a line that cuts across the curve.

The tangent DOES NOT cut across the curve! It only just touches it at one point.)  

Derivative: The derivative of a function is another function that gives you its gradient. For example, take the function y = x2. Its derivative is 2x, so when x = 2, the gradient of the curve is 4; when x = 6, the gradient of the curve is 12, and so on. Normal: The normal is a line perpendicular to the tangent that crosses the curve at the same

point: 

Polynomial: An equation that just has powers of x or constants in it. ◦ y = x, y = x5, y = 4x-5 +7x6 + 3x, y = 3 are all polynomials. ◦ y = sin x, y = ex, y = 2x are not polynomials. ◦ At AS, differentiation focuses on polynomials; you'll learn how to differentiate other functions in C3.

also that there are different ways to write the derivative. y = some function of x -----> dy/dx = its derivative f(x) = some function of x ----> f'(x) = its derivative (pronounced f-prime x) Also note that it doesn't have to be y and x. For example: if y = x2 then dy/dx = 2x but if t = q2 then dt/dq = 2q. The method is exactly the same as for y and x, only the variables have different letters. Make sure you'd write dt/dq for the derivative and not dy/dx.

Finding the derivative 

For the equation y = 2, y = -4, y = 6 or y = any constant number, the graph will be a horizontal, flat straight line:

In this case it's pretty obvious that dy/dx = 0, since the graph is flat and doesn't slope at all! e.g. y = 7, therefore dy/dx = 0.  For a straight line graph of the form y = mx + c, e.g. y = x, y = 3x + 7, y = 8x etc. the gradient is just m, the coefficient of x. e.g. y = 5x + 7, therefore dy/dx = 5.  For a power of x, e.g. y = 4x2, y = 6x7, etc, of the form y = axb: dy/dx = abxb-1. Rather than ing the formula though I just find it easier to "take one off the power and multiply the coefficient by the old power". e.g. y = 6x4, therefore dy/dx = (6 times 4)x(4-1) = 24x3. y = 4x5, therefore dy/dx = (4 times 5)x(5-1) = 20x4. y = x8, therefore dy/dx = (8 times 1)x(8-1) = 8x7.  For a polynomial equation, you differentiate each term seperately. For example: y = 7x6 + 3x2 + 6x + 3 dy/dx = 42x5 + 6x + 6 (+0)

 >>> THINGS TO HERE! <<< ◦ your power rules: x-2 means 1/x2, x1/2 means the square root of x, x3/2 means (the square root of x)3, etc. If you're rusty on those then ask me, because you might get a question asking you to differentiate an alternative form, e.g. the square root of x rather than x1/2, or 1/x4 rather than x-4. ◦ If you get brackets, expand them out first. That should give you a polynomial that you

can easily differentiate using the rule above. ◦ Be careful with negative numbers! Say you want to differentiate x- 4. Your first instinct might be to think the answer is -4x-3, but that -4 minus one is -5, so the answer is -4x-5! ◦ If you get a constant, its derivative is zero. You don't need to include that in your answer, because it doesn't affect anything. Also, that for a power term such as x5 the coefficient of x is 1.

Worked examples:

1a) First we want to get this into a polynomial form so we can differentiate it. from your power rules that the square root of x = x1/2. So: y = x4 + 6√x y = x4 + 6x1/2 We then use the rules for polynomial differentiation mentioned earlier to find dy/dx:  When differentiating a polynomial, you differentiate each term separately  When differentiating something of the form axb, dy/dx = abxb-1. therefore dy/dx = 4x3 + 3x-1/2. b) First we should expand the brackets: (x+4)2 = (x+4)(x+4) = x2 + 8x + 16 y = (x+4)2/x, so: y = (x2 + 8x + 16)/x = x + 8 + 16/x (Hint: if asked to divide a polynomial by x, divide every term by x.) We now put this in polynomial form:

y = x + 8 + 16x-1 And differentiate: dy/dx = 1 - 16x-2 Finding equations of tangent lines that by definition, the derivative will give you the gradient of the tangent line. Also that the equation of a straight line is y = mx + c, and that m is the gradient. To find the equation of a tangent to a curve at a point (x, y), all you have to do is: 1. Differentiate the equation of the curve, to get the gradient function, 2. Plug x into the derivative to find out the value of the gradient at that point, 3. Write down y = mx + c, with m as your value of the gradient, and 4. Plug in the values of x and y into that equation to find c. Here's a worked example of this:

a) First, expand the brackets: (x+3)(x-8) = x2 + 3x - 8x - 24 = x2 - 5x - 24 To find y we should divide all this by x: y = x - 5 - 24/x Put in polynomial form: y = x - 5 - 24x-1

And differentiate: dy/dx = 1 + 24x-2. (or dy/dx = 1 + 24/x2.) b) We're asked to find the tangent at the point x = 2, so plug x = 2 into the derivative: dy/dx = 1 + 24/x2 Therefore dy/dx = 1+24/22 = 1 + 24/4 = 1+6 = 7. So this is the gradient of the tangent at x=2. The tangent is a straight line with equation y = mx + c, and since m is the gradient: y = 7x + c. We need to find the value of y, so plug x back into the original equation to find out what y is when x=2: y = [(x-3)(x-8)]/x = [(2-3) (2-8)]/2 = (-1 times -6) /2 = 6/2 = 3. So now we know x=2 and y=3, so plug these values into our equation for the tangent line: y = 7x + c therefore 3 = 7 times 2 + c = 14 + c Therefore c = -11. Therefore y = 7x - 11. Equations of normals and parallel lines A normal is a line perpendicular to the tangent that crosses the curve at the same point as the tangent. The important thing to here is that if two lines are perpendicular, then if the gradient of one is m, the gradient of the other will be -1/m. (We call this the negative reciprocal of m.) You can use this fact to help you find equations of normals at a point. Here is a worked example:

We know y = 3x2 - 8x + 7 This is a nice polynomial so we can differentiate it easily.

dy/dx = 6x - 8. We know the x-co-ordinate of P is 2, so x=2, so we plug x=2 into the derivative: dy/dx = 6x - 8 = (6 times 2) - 8 = 12-8 = 4. Since the gradient of the tangent is 4, and the normal is perpendicular to the tangent, the gradient of the normal must be -1/4. The normal is a straight line, so its equation is of the form y = mx + c, so: y = -1/4x + c We do the same thing we did in the tangent example. Because the normal crosses the curve at point P, we can plug x=2 into the curve's original equation to find y: y = 3x2 - 8x + 7 y = 3(2)2 - 8 times 2 + 7 = 3 times 4 - 16 + 7 = 12 - 16 + 7 = 3. So we know x = 2 and y = 3, so plug that into our equation for the normal: y = -1/4x + c therefore 3 = -1/4 times 2 + c therefore c = 3.5. Therefore y = -1/4x + 3.5. Another important thing to is that if two lines are parallel, their gradients will be the same. This means that if you differentiate them, at the same point you should get the same value of the derivative. 

>>> CHECKLIST <<<

Do you understand the meaning of the words: function, gradient, tangent, normal, derivative, polynomial, negative reciprocal?  Can you differentiate any of the functions mentioned above, including constants (such as y = 3), in x (such as y = 6x), in powers of x (such as y = 7x6) and polynomials (such as y = 6x3 + 7x-5 + 3)?  Can you use differentiation to find the gradient of the tangent to a graph at a point, and then use y = mx + c to find the equation of the tangent line at that point?  Can you use the negative reciprocal of the gradient to find the equation of the normal to a curve at any point? >>> EXAM HINTS TO <<  Always expand out brackets first!  your power rules, so you can convert functions like √x and 6/x6 into polynomials!  If you're dividing a polynomial by something (e.g. by x), you divide every term by it!  If two lines are perpendicular, the gradient of one will be the negative reciprocal of the other!  If two lines are parallel, they will have the same gradient! What now? Here's another worked example for you to try:

Let me know if you have any problems with it! Here are some useful web links on differentiation in the mean time: Differentiation: http://www.mathsrevision.net/alevel/pages.php?page=9 Tangents and normals: http://www.mathsrevision.net/alevel/pages.php?page=41 S-cool differentiation: http://www.s-cool.co.uk/a-level/maths/differentiation (Note: this goes beyond C1 into C2 and C3 territory) Some videos: Khan Academy on derivatives: http://www.youtube.com/watch?v=rAof9Ld5sOg Actual C1 worked examples!: http://www.youtube.com/watch?v=mUJDUTY6iU4 Tangents and normals: http://www.youtube.com/watch?v=Iu67T-NcXvc&feature=related Good luck with your revision!