Revision:OCR Core 1 - Differentiation

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1 5. Differentiation
2 Also See
3 Comments

5. Differentiation

The gradient of a curve

It was shown earlier in these notes that it is simple to find the gradient of a line, but curves are different. Consider the definition of gradient that was given previously:

$m = \frac{\Delta y}{\Delta x}$

On a curve this is different for different points. For instance, one can consider the curve of $y = x^{2}$ .

$When \ x = 1, \ y = 1$

$When \ x = 2, \ y = 4$

$When \ x = 3, \ y = 9$

$When \ x = 4, \ y = 16$

Consider the gradient, m, between these points:

$m = \frac{\Delta y}{\Delta x} = \frac{4 - 1}{2 - 1} = 3$

$m = \frac{\Delta y}{\Delta x} = \frac{9 - 4}{3 - 2} = 5$

$m = \frac{\Delta y}{\Delta x} = \frac{16 - 9}{4 - 3} = 7$

This shows that the gradient is changing.

One would define the gradient of a curve at a point, P, to be the gradient of the tangent to the curve at this point (the tangent being a line which touches the curve in only one place).

How would one calculate this?

If one knew the equation of the tangent, one could calculate the gradient of the curve at the point of contact, but otherwise the previous formula does not work (as it relies upon two values for x, and y; and as there is merely one point of contact, there is a single value of x, and a single value of y).

Chords are lines which intersect the curve at more than one place, with these straight lines, one can give an approximation of the gradient of the curve at a point.

One might deduce that as one allows the two points of intersection on the curve (of the chord) to become closer, the approximation of the gradient of the curve at a point will become more and more accurate (with the definition of the point as that point which the limit of the two intersections of the curve is).

What one might consider is the chord between a given point (on the curve) and the point at which the gradient of the curve is what one wishes to calculate. As the point tends towards the point at which the gradient will be calculated, the approximation becomes better.

Consider the diagram below.

Figure 004

One wishes to calculate the gradient at the point (1, 1) on the above curve ( $y = x^{2}$ ).

Now consider values of x which are close to the point (1, 1) (as one knows that as one tends towards 1, the gradient of the chord will tend towards the gradient of the tangent to the curve at (1, 1)).

$When \ x = 1, \ y = 1$

$When \ x = 1.1 \ y = 1.21$

$When \ x = 1.01, \ y = 1.0201$

$When \ x = 1.001, \ y = 1.002001$

Now consider the gradients of each of the chords:

$m = \frac{\Delta y}{\Delta x} = \frac{1.21 - 1}{1.1 - 1} = 2.1$

$m = \frac{\Delta y}{\Delta x} = \frac{1.0201 - 1}{1.01 - 1} = 2.01$

$m = \frac{\Delta y}{\Delta x} = \frac{1.002001 - 1}{1.001 - 1} = 2.001$

A pattern emerges, and it seems that the gradient is tending towards 2.

Gradient formulae

One can consider the gradient at several different points on the curve (in Figure 004) in order to get an understanding of the relationship between the gradient (an estimate at this stage), and the point at which this gradient is taken from.

Consider the following:

$When \ x = 2, \ y = 4$

$When \ x = 2.1 \ y = 4.41$

$When \ x = 2.01, \ y = 4.0401$

$When \ x = 2.001, \ y = 4.004001$

Now consider the gradients of each of the chords:

$m = \frac{\Delta y}{\Delta x} = \frac{4.41 - 4}{2.1 - 2} = 4.1$

$m = \frac{\Delta y}{\Delta x} = \frac{4.0401 - 4}{2.01 - 2} = 4.01$

$m = \frac{\Delta y}{\Delta x} = \frac{4.004001 - 4}{2.001 - 2} = 4.001$

This seems to tend towards 4.

Now consider:

$When \ x = 3, \ y = 9$

$When \ x = 3.1 \ y = 9.61$

$When \ x = 3.01, \ y = 9.0601$

$When \ x = 3.001, \ y = 9.006001$

Now consider the gradients of each of the chords:

$m = \frac{\Delta y}{\Delta x} = \frac{9.61 - 9}{3.1 - 3} = 6.1$

$m = \frac{\Delta y}{\Delta x} = \frac{9.0601 - 9}{3.01 - 3} = 6.01$

$m = \frac{\Delta y}{\Delta x} = \frac{9.006001 - 9}{3.001 - 3} = 6.001$

This seems to tend towards 6.

One might be able to deduce something.

Consider that if x = 1, the estimate for the gradient is 2; if x = 2, the estimate for the gradient is 4; and if x = 3, the estimate for the gradient is 6. This seems to suggest that the gradient at any point on the curve is found by .

One might say that the gradient formula for $y = x^{2}$ was .

The process which has been carried out is differentiation; one has differentiated the function of x $f(x) = x^{2}$ .

There is a notation based upon function notation for this differentiated version of the function of x.

$f^{'}(x)$

This is "f dashed x", hence:

$f(x) = x^{2}$

$f^{'}(x) = 2x$

Some rules for differentiation

For terms of $x^{2}$ , when differentiated, they become a multiple of .

For terms which are of , when differentiated, they become a constant value.

Constant values differentiate to nothing, which is perfectly logical. If one thinks back to the effect of the constant term on the general quadratic expression ( $y = ax^{2} + bx + c$ ) it is merely a move in the y-direction. This does not change the gradient, and hence any function whose a, and b coefficients are identical, but have different constant terms will have the same gradient.

Extending the differentiation rule

One has seen that one can obtain a rule for the differentiation of terms of $x^{2}$ , but this is only the beginning, one needs to understand the rules for terms of higher order.

It is useful to be able to generalise somewhat.

Consider the curve:

This is a generic function of x.

One was calculating a succession of gradients of chords who had a common point (the point at which one wished to obtain the gradient), and then observing tendencies.

Suppose one wishes to find the gradient at the point: $(x, \ f(x))$ .

Consider a point slight further along the curve: $(x + \delta x, \ f(x + \delta x))$ .

To find the gradient one uses the following:

$m = \frac{\Delta y}{\Delta x} = \frac{f(x + \delta x) - f(x)}{x + \delta x - x} = \frac{f(x + \delta x) - f(x)}{\delta x}$

Now consider the effects of the following:

$\lim_{\delta x \to 0} \frac{f(x + \delta x) - f(x)}{\delta x}$

What one is observing is an algebraic representation of taking the gradient of a chord between a point that is becoming infinitely close to the point at which one wishes to calculate the gradient, and the point that one wishes to calculate the gradient of the curve at itself.

Now that we define the function of x:

$f(x) = x^{3}$

Hence, one would do the following:

$f^{'}(x) = \lim_{\delta x \to 0} \frac{f(x + \delta x) - f(x)}{\delta x} = \lim_{\delta x \to 0} \frac{(x + \delta x)^{3} - x^{3}}{\delta x}$

Hence:

$f^{'}(x) = \lim_{\delta x \to 0} \frac{x^{3} + 3x^{2} \delta x + 3x (\delta x)^{2} + (\delta x)^{3} - x^{3}}{\delta x} = \lim_{\delta x \to 0} \left[ 3x^{2} + \frac{3x \delta x + (\delta x)^{2}}{1} \right]$

Hence:

$f^{'}(x) = 3x^{2}$

Now, one can make the function slightly more generic:

$f(x) = x^{n}$

Hence one can calculate:

$f^{'}(x) = \lim_{\delta x \to 0} \frac{f(x + \delta x) - f(x)}{\delta x} = \lim_{\delta x \to 0} \frac{x^{n} + {n \choose 1} x^{n - 1} \delta x + {n \choose 2} x^{n - 2} (\delta x)^{2} + ... + (\delta x)^{n} - x^{n}}{\delta x}$

Hence:

$f^{'}(x) = \lim_{\delta x \to 0} \left[ {n \choose 1} x^{n - 1} + \left( \frac{{n \choose 2} x^{n - 2} \delta x + {n \choose 3} x^{n - 3} (\delta x)^{2} + ... + (\delta x)^{n - 1}}{1} \right) \right] = {n \choose 1} x^{n - 1}$

Hence:

$f^{'}(x) = nx^{n - 1}$

This leads to the general rule:

$f(x) = x^{n}$

Hence:

$f^{'}(x) = nx^{n - 1}$

It is also important to know another few rules.

Consider:

Hence:

$h^{'}(x) = \lim_{\delta x \to 0} \frac{h(x + \delta x) - h(x)}{\delta x}$

$f^{'}(x) = \lim_{\delta x \to 0} \frac{f(x + \delta x) - f(x)}{\delta x}$

$g^{'}(x) = \lim_{\delta x \to 0} \frac{g(x + \delta x) - g(x)}{\delta x}$

Hence:

$f^{'}(x) + g^{'}(x) = \lim_{\delta x \to 0} \frac{f(x + \delta x) - f(x) + g(x + \delta x) - g(x)}{\delta x}$

$\lim_{\delta x \to 0} \frac{f(x + \delta x) - f(x) + g(x + \delta x) - g(x)}{\delta x} = \lim_{\delta x \to 0} \frac{h(x + \delta x) - h(x)}{\delta x}$

Hence:

$h^{'}(x) = f^{'}(x) + g^{'}(x)$

Also, one might consider:

$a \times f(x)$

Now, consider the derivative:

$\lim_{\delta x \to 0} \frac{a \times f(x + \delta x) - a \times f(x)}{\delta x}$

One can take a factor of "a", giving:

$\lim_{\delta x \to 0} a \left[ \frac{f(x + \delta x) - f(x)}{\delta x} \right]$

Hence, one can say that the derivative of a function which is multiplied by a constant, is that constant value multiplied by the derivative of the function.

One can use these rules to aid in the calculation of a derivative.

Example

1. Calculate the derivative of the function of x:

$f(x) = 3x^{2} + 2x + 1$

First one can express this as:

$f(x) = 3 \times g(x) + 2 \times h(x) + 1$

Hence (as a constant value has a derivative of zero):

$f^{'}(x) = 3 \times g^{'}(x) + 2 \times h^{'}(x)$

Hence:

$3 \times g^{'}x = 3 \times (2x)$

$2 \times h^{'}x = 2 \times (1)$

Hence:

$f^{'}(x) = 3 \times (2x) + 2 \times (1) = 6x + 2$

An alternative notation

Functional notation is helpful in some situations, but much like other forms of notation, it is not necessarily the best notation to use in all circumstances.

If one is referring to a graph where one is graphing a function of a variable to another variable, using the usual notation is better (note, the use of functional notation in the previous function is merely for the correct representation of the notation, it is not to do with functional notation as the reader should notice as they read on).

It is noteworthy at this point that the gradient of a chord can be expressed as:

$\frac{\Delta y}{\Delta x}$

or,

$\frac{\delta y}{\delta x}$

(Both are equivalent).

This is not the notation for the derivative (this is a mistake that some people make).

In order to differentiate between this gradient of a chord, and the gradient of the curve at a single point, one uses the following notation for the derivative:

$\frac{dy}{dx}$ .

This notation is read as "d y d x" (it is not a fraction).

One should also note that one cannot simply say that "the d y d x of $x^{2}$ is ". One must refer to a function of x that gives a value to y. Hence, one can say:

" $y = x^{2}$ , hence d y d x equals ".

If one wishes to use this style of notation, but it is not convenient to use "y", one can put:

$\frac{d}{dx} \left( x^{2} \right) = 2x$ .

Finding the equation of tangents

If one is aware of the method for calculating the equation of a straight line, and one is also aware that when one uses differentiation to obtain a value for the gradient of the curve at a given point, one is in fact finding the gradient of the tangent which has a point of contact which is the point at which one wished to find the gradient of; one can find the equation of the tangent to the curve at the point which one knows the gradient of the curve at.

It is simple to see that if one is aware of the gradient of the tangent to a curve (at a given point), and one is aware that the tangent is a straight line, one can use the method that one has already studied in order to calculate the equation of the tangent.

Example

1. Calculate the equation of the tangent to the curve $y = x^{2}$ at the point where .

This is a relatively simple problem, it involves three stages: initially, one must find the derivative of the equation, then one must use this to find the gradient at the given point, then one must calculate the equation of the tangent.

First, the derivative is found:

$\frac{dy}{dx} = 2x$

Now, find the gradient:

$m = 2x = 2 \times 20 = 40$

Now use the general form of the equation of a straight line:

One will need to know a point upon the line, and one knows that when , the tangent does indeed touch the curve (hence this point must be on the tangent). To calculate the point, one can use the original function of x:

$y = x^{2} = 20^{2} = 400$

Hence the point of contact is (20, 400).

Now, one can go back to the equation of the tangent:

Now one can substitute:

Hence:

$400 = 40 \times 20 + c = 800 + c$

Hence:

Hence, the equation of the tangent is:

The normal to a curve at a point

The normal to a curve at a point is merely the formal way to define the perpendicular to the tangent of the curve at a point.

Calculations surrounding this are all explained in depth in the first section of these notes, and thus it is advisable for those who do not feel that they fully understand the example whose succession of this text is imminent, should re-read (or read, in fact) the text in the aforementioned second (or part of section).

Example

1. Calculate the equation of the normal to the curve $y = x^{2}$ at the point where .

First one must calculate the derivative of the function such that one is able to obtain a numerical value for the gradient of the tangent to the curve at the given point, hence:

$\frac{dy}{dx} = 2x$

Hence, the gradient of the tangent:

$2x = 2 \times 4 = 8$

One can now calculate the gradient of the perpendicular to this tangent:

$m = 8, \ - \frac{1}{m} = - \frac{1}{8}$

Now apply the general formula for the equation of a straight line:

Hence (the change from "c" to "k" is to avoid confusion when one changes the form of the equation):

$y = - \frac{1}{8}x + k$

$Let \ 8k = c$

Hence one should find the coordinates of the point of contact of the tangent with the curve (as the normal to the tangent at this point will, by definition, go through the point).

Hence:

$y = x^{2} = 4^{2} = 16$

Hence the point is (4, 16).

Now one can substitute:

$8 \times 16 + 4 = c = 132$

Hence the equation of the normal to the curve at the point at which is:

Also See

Read these other OCR Core 1 notes:

Comments

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Article Updates

Revision:OCR Core 1 - Differentiation

Contents

5. Differentiation

The gradient of a curve

Gradient formulae

Some rules for differentiation

Extending the differentiation rule

An alternative notation

Finding the equation of tangents

The normal to a curve at a point

Also See

Comments