CHAIN RULE OR DIFFERENTIATING A FUNCTION OF A FUNCTION

5. CHAIN RULE OR DIFFERENTIATING A FUNCTION OF A FUNCTION

You already know about the chain rule whether you have called it that or not. Look at this example.

Example 1. Find dy/dx if y = ln sin 2x. You would say dy

(2x) = 2 cot 2x. dx

(sin 2x) =

We could write this problem as y = ln u,

v = 2x. Then we would say

This is an example of the chain rule. We shall want a similar equation for a function of several variables. Consider another example.

200 Partial Differentiation Chapter 4

Example 2. Find dz/dt if z = 2t 2 sin t.

Differentiating the product, we get

We could have written this problem as z = xy,

where x = 2t 2 and y = sin t, dz

But since x is ∂z/∂y and y is ∂z/∂x, we could also write

We would like to be sure that (5.1) is a correct formula in general, when we are given any function z(x, y) with continuous partial derivatives and x and y are differentiable functions of t. To see this, recall from our discussion of differentials that we had

where ǫ 1 and ǫ 2 → 0 with ∆x and ∆y. Divide this equation by ∆t and let ∆t → 0; since ∆x and ∆y → 0, ǫ 1 and ǫ 2 → 0 also, and we get (5.1). It is often convenient to use differentials rather than derivatives as in (5.1). We would like to be able to use (3.6), but in (3.6) x and y were independent variables and now they are functions of t. However, it is possible to show (Problem 8) that dz as defined in (3.6) is a good approximation to ∆z even though x and y are related. We may then write

whether or not x and y are independent variables, and we may think of getting (5.1) by dividing (5.3) by dt. This is very convenient in doing problems. Thus we could do Example 2 in the following way:

dz = x dy + y dx = x cos t dt + y · 4t dt = (2t 2 cos t + 4t sin t)dt, dz

= 2t 2 cos t + 4t sin t. dt

In doing problems, we may then use either differentials or derivatives. Here is another example.

Section 5 Chain Rule or Differentiating a Function of a Function 201 Example 3. Find dz/dt given z = x y , where y = tan −1 t, x = sin t.

Using differentials, we find dt

2 dt . 1+t You may wonder in a problem like this why we don’t just substitute x and y

= yx y−1 cos t + x y

ln x ·

as functions of t into z = x y to get z as a function of t and then differentiate. Sometimes this may be the best thing to do but not always. For example, the resulting formula may be very complicated and it may save a lot of algebra to use (5.1) or (5.3). This is especially true if we want dz/dt for a numerical value of t. Then there are cases when we cannot substitute; for example, if x as a function of t is given by x+ e x = t, we cannot solve for x as a function of t in terms of elementary functions. But we can find dx/dt, and so we can find dz/dt by (5.1). Finding dx/dt from such an equation is called implicit differentiation; we shall discuss this process in the next section.

Computers can find derivatives, so why should we learn the methods shown here and in the following sections? Perhaps the most important reason is that the techniques are needed in theoretical derivations. However, there is also a practical reason: When a problem involves a number of variables, there may be many ways

to express the answer. (You might like to verify that dz/dt = z(y cot t + ln x cos 2 y) is another form for the answer in Example 3 above). Your computer may not give you the form you want and it may be as easy to do the problem by hand as to convert the computer result. But in problems involving a lot of algebra, a computer can save time, so a good study method is to do problems both by hand and by computer and compare results.

PROBLEMS, SECTION 5

1. Given z = xe −y , x = cosh t, y = cos t, find dz/dt. √

2. Given w = u 2 +v 2 , u = cos[ln tan(p + 1 4 1 π)], v = sin[ln tan(p + 4 π)], find dw/dp. −p 2 −q 3. 2 Given r = e ,p=e s ,q=e −s , find dr/ds.

4. Given x = ln(u 2 −v 2 ), u = t 2 , v = cos t, find dx/dt.

5. If we are given z = z(x, y) and y = y(x), show that the chain rule (5.1) gives

6. Given z = (x + y) 5 , y = sin 10x, find dz/dx.

7. Given c = sin(a − b), b = ae 2a , find dc/da. 8. Prove the statement just after (5.2), that dz given by (3.6) is a good approximation

to ∆z even though dx and dy are not independent. Hint: let x and y be functions independent variables). However, ∆x/∆t is nearly dx/dt for small dt and dt = ∆t

since t is the independent variable. You can then show that

„ dx «

∆x =

+ǫ x dt = dx + ǫ x dt,

dt

202 Partial Differentiation Chapter 4

and a similar formula for ∆y , and get

∂z

∂z

∂y dy + (terms containing ǫ’s) · dt = dz + ǫdt, where ǫ → 0 as ∆t → 0.