Revision:OCR Core 3 - The chain rule - The Student Room
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9. The chain rule

The chain rule: and informal treatment

Earlier in these notes the idea of differentiating a function of the form  (ax + b)^{n} was discussed.

This function can be thought of as a function of a function, in the sense that there is a function, "f", such that  f(x) = ax + b , and a function, "g", such that  g(f(x)) = (ax + b)^{n} .

The chain rule is an attempt to generalise this such that it is possible to differentiate a function in the form:  f_{n}f_{n - 1} ... f_{2}f_{1}(x) (where each function is differentiable).

The idea of the chain rule lies in the delegation of a statement to a letter.

Consider the example:  \frac{d}{dx} (ax + b)^{n} = na(ax + b)^{n - 1} .

In this case a candidate might naturally think to let  u = ax + b , hence the problem becomes:  \frac{d}{dx} u^{n} .

Notice the result which is known (from the previous notes), and the derivative of this new component.

It seems that if  y = (ax + b)^{n}, \ and \ u = ax + b ,  \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} .

(Note that if the derivatives were fractions this would be true [this is often a good memory technique], however, the derivatives should not be thought of as fractions [the notation is coincidental]).

Considering this process as rates of change can also be interesting.

The change in "y" with respect to "x" is  \frac{dy}{dx} .

The change in "y" with respect to "u" is  \frac{dy}{du} .

The change in "u" with respect to "x" is  \frac{du}{dx} .

Immediately it seems that the aforementioned connection should be drawn.

To more formally state the definition:

 y = f_{n}f_{n - 1} ... f_{2}f_{1}(x)

 u_{t} = f_{t}(x)  \implies \frac{dy}{dx} = \frac{dy}{du_{n}} \times \frac{du_{n}}{du_{n - 1}} \times ... \times \frac{du_{2}}{du_{1}} \times \frac{du_{1}}{dx} .


Deriving the chain rule

The idea (met in Core 1) of  \frac{dy}{dx} was the instantaneous change in "y" with respect to "x". The idea of  \frac{ \delta y }{ \delta x } was used as a precursor; the average change in "y" with respect to "x", representing the gradient of a secant line. Upon limitation, the following is defined:  \frac{dy}{dx} = \lim _{ \delta x \to 0 } \frac{ \delta y }{ \delta x } , as the closer the secant line gets to a point, the closer it gets to the same gradient as the point.

The treatment of the average (using delta) can be one of a fraction, as there is a division, hence:

 \frac{ \delta y }{ \delta x } = \frac{ \delta y }{ \delta u } \times \frac{ \delta u }{ \delta x } .

This arises from a small increase in "x" resulting in a corresponding increase in "u", and finally an increase in "y".

Using limits the derivative can be found:

 \frac{dy}{dx} = \lim _{ \delta x \to 0 } \frac{ \delta y }{ \delta x } = \lim _{ \delta x \to 0 } \left( \frac{ \delta y }{ \delta u } \times \frac{ \delta u }{ \delta x } \right) .

Two assumptions are made at this level;  \delta x \to 0 \implies \delta u \to 0 , and  \lim _{ \delta x \to 0 } \left( \frac{ \delta y }{ \delta u } \times \frac{ \delta u }{ \delta x } \right) = \lim _{ \delta x \to 0 } \left( \frac{ \delta y }{ \delta u } \right) \times \lim _{ \delta x \to 0 } \left( \frac{ \delta u }{ \delta x } \right) .

Hence:

 \frac{dy}{dx} = \lim_{ \delta u \to 0 } \left( \frac{ \delta y }{ \delta u } \right) \times \lim _{ \delta x \to 0 } \left( \frac{ \delta u }{ \delta x } \right)

Hence:

 \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} .

An application of the chain rule

The chain rule can be used to extend the already demonstrated (in Core 1) rule that  \frac{d}{dx} x^{n} = nx^{n - 1} from  n \in \mathbb{Z} to  n \in \mathbb{R} .

 x^{n} = e^{ \ln \left( x^{n} \right) } = e^{n \ln x } .

 u = n \ln x

 \frac{d}{dx} x^{n} = e^{u} \times \frac{n}{x}

Hence:

 \frac{d}{dx} x^{n} = e^{n \ln x} \times \frac{n}{x} = nx^{n - 1} .

Related rates of change

In many cases the relationship between area, volume and time can enable a candidate to use the chain rule to deduce the rate of change of a quantity.

Example

1. Ripples in a pond are such that they form perfect circles, the outer-most of which has a radius that changes at a rate of  5 \ cms^{-1} ; calculate the rate at which the areachange at the time when the radius is 100cm.

Consider that  A = \pi r^{2} , and  \frac{dr}{dt} = 5 .

 \frac{dA}{dt} = \frac{dA}{dr} \times \frac{dr}{dt} = \frac{dA}{dr} \times 5

 \frac{dA}{dr} = 2 \pi r = 200 \pi (As the radius is known).

Hence:

 \frac{dA}{dt} = 1000 \pi \ cm^{2}s^{-1} .

Be sure to use the correct units (and quote them in the answer).

The relation between  \frac{dy}{dx} , and  \frac{dx}{dy}

Sometimes it is useful to know what the rate of change of "x" with respect to "y". One example is when the candidate wishes to calculate the points at which the gradient of a curve is parallel to the y-axis, as this would imply an undefined rate of change of "y" with respect to "x", but a zero rate of change of "x" with respect to "y".

It is useful to consider a simple example:

 y = mx + c

 x = \frac{y - c}{m}

Hence:

 \frac{dy}{dx} = m

 \frac{dx}{dy} = \frac{1}{m} .

It therefore seems that the relationship is:  \displaystyle \frac{dx}{dy} = \frac{1}{ \frac{dy}{dx} } .

Consider the very basic technique of finding the gradient of a curve at a point, draw a tangent, and calculate the gradient of the straight line.

If there is a graph of  y = f(x) , the gradient of the tangent at a point is equal to the tangent of the angle that this tangent makes with the x-axis.

Consider a function of "y":  x = g(y) .

Following a convention, the "y" would be on the bottom axis (as it is independent), and the "x" would go upwards.

Similarly, the angle which the tangent at a point makes with the x-axis produces the gradient of the tangent when the tangent function is applied to it.

Assuming that the functions have been chosen such that  y = f(x) \Leftrightarrow x =g(y) , the axes of the graph of  x = g(y) can be switched around to form what is the same graph as  y = f(x) .

The angles written of earlier now will add to make  \frac{ \pi }{2} .

Hence, the tangent of one of these (the gradient of the tangent at the point) is equal to the cotangent of the other. Therefore the relationship is true for the cases when  y = f(x) \Leftrightarrow x = g(y) .

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