Handout 5, Summer 2014 Math May Consider the following table of values: x f(x) g(x) f (x) g (x)

Transcription

1 Handout 5, Summer 204 Math May 204. Consider the following table of values: x f(x) g(x) f (x) g (x) Let h(x) = (f g)(x) and l(x) = g(f(x)). Compute h (3), h (4), l (8), and l (3). Solution. Since h(x) = f(g(x)), by the chain rule we have that h (x) = f (g(x)) g (x), so that we have: and h (3) = f (g(3))g (3) = f (8) = 9 = 99 h (4) = f (g(4))g (4) = f (4) 9 = 2 9 = 8 Next, since l(x) = g(f(x)), we again have by the chain rule that l (x) = g (f(x))f (x). Therefore, and We have the result. l (8) = g (f(8))f (8) = g (8) 9 = 4 9 = 36 l (3) = g (f(3))f (3) = g (4) = 9 2. Suppose g(x) = f(x). Given the graph of f below, calculate g (3).

2 2 Solution. Note that g(x) is the composition of the functions o(x) = x and i(x) = f(x). Therefore, using the chain rule, we have o(x) = x i(x) = f(x) o (x) = 2 x /2 i (x) = f (x) Thus, g (x) = (o(i(x))) = o (i(x))i (x) = 2 (f(x)) /2 f (x). This gives g (3) = f (3). From the graph, we can immediately see that f(3) = 2. 2 f(3) Furthermore, since the derivative of the graph at a point is the slope of the line tangent to the graph at that point, we know that f (x) = 2. 3 Indeed, the slope of the blue line in the graph is 2. Therefore, we 3 have: g (3) = f (3) 2 f(3) = = 3 2 This gives the result. 3. If g is twice differentiable (i.e. has a first and second derivative) and f(x) = xg(x 2 ), find f (x). Solution. Here we have a product of two functions: x and g(x 2 ). Therefore, we use the product rule first to take the derivative of f: f (x) = x g(x 2 ) + x(g(x 2 )) = g(x 2 ) + x(g(x 2 )) Now, g(x 2 ) is a composition of functions: o(x) = g(x) i(x) = x 2 o (x) = g (x) i (x) = 2x Therefore, using the chain rule, we have that (g(x 2 )) = o (i(x))i (x) = 2xg (x 2 ). Thus, we have f (x) = g(x 2 ) + 2x 2 g (x 2 ) To find f (x), we follow a similar procedure. We have: This gives the result. f (x) = 2xg (x 2 ) + 4xg (x 2 ) + 4x 3 g (x 2 )

3 4. Find d03 03 cos(2x) by computing the first few derivatives and observing a pattern. Solution. Let s begin with the first derivative of cos(2x). composition of functions: o(x) = cos(x) i(x) = 2x o (x) = sin(x) i (x) = 2 3 This is a By the chain rule, then, we have that (cos(2x)) = 2 sin(2x). Similarly, we can compute the next few derivatives: (cos(2x)) = 2 sin(2x) (cos(2x)) = 4 cos(2x) (cos(2x)) = 8 sin(2x) (cos(2x)) (4) = 6 cos(2x) And so on. Note that every fourth derivative is some multiple of cos(2x). Furthermore, the coefficient of the derivative is ± a power of 2. Putting this together, we can see that (cos(2x)) (00) = 2 00 cos(2x). Therefore, we have: We have the result. (cos(2x)) (0) = 2 0 sin(2x) (cos(2x)) (02) = 2 02 cos(2x) (cos(2x)) (03) = 2 03 sin(2x) 5. Calculate dy if 2 x + y = 3. Solution. Replacing y with f(x), we have: 2 x + f(x) = 3 Differentiating both sides with respect to x, we obtain: ( ) 2 2 x /2 + 2 f(x) /2 f (x) = 0 That is, + f (x) x 2 f(x) = 0

4 4 We now solve for f (x). We have: + f (x) x 2 f(x) = 0 f (x) 2 f(x) = x f (x) = 2 f(x) x Finally, swapping all f(x) for y and all f (x) for y, we arrive at: dy = y = 2 y y = 2 x x 6. Calculate dy if 2x3 + x 2 y xy 3 = 2. Solution. Here we will proceed without swapping y for f(x). See if you can follow along. We differentiate both sides with respect to x. Note that a product rule is needed on the second term of the left side, and a product and chain rule are needed on the third term of the left side. We have: We have the result. 6x 2 + (2xy + x 2 y ) (y 3 + 3xy 2 y ) = 0 6x 2 + 2xy + x 2 y y 3 3xy 2 y = 0 6x 2 + 2xy y 3 = 3xy 2 y x 2 y 6x 2 + 2xy y 3 = y (3xy 2 x 2 ) y = dy = 6x2 + 2xy y 3 3xy 2 x 2 7. Calculate dy if 4 cos(x) sin(y) =. Solution. Here we go into more explicit detail. First, let s swap y for f(x). Then we have 4 cos(x) sin(f(x)) = To differentiate both sides, we must use a product and chain rule on the left-hand side. That is, by the product rule we have: 4( sin(x)) sin(f(x)) + 4 cos(x)(sin(f(x))) = 0

5 Finding (sin(f(x))) and plugging it into this equation will complete the differentiation. To compute this, we must use the chain rule. o(x) = sin(x) i(x) = f(x) o (x) = cos(x) i (x) = f (x) Therefore, by the chain rule (sin(f(x))) = cos(f(x))f (x). Plugging this into the equation above, we have 4 sin(x) sin(f(x)) + 4 cos(x) cos(f(x))f (x) = 0 At last, solving for f (x), we have: f (x) = 4 sin(x) sin(f(x)) 4 cos(x) cos(f(x)) Therefore, swapping f(x) back for y and f (x) back for y, we arrive at our final answer: y = dy sin(x) sin(y) = cos(x) cos(y) 5 8. Calculate dy if x + y = + x 2 y 2. Solution. Again, we will work out this result in detail. We begin by replacing y by f(x). We have: x + f(x) = + x 2 f(x) 2 To find dy, we need to start by differentiating both sides. To accomplish this, we must use a chain rule on the left-hand side, and a product and chain rule on the right-hand side. For the left-hand side, we have: o(x) = x i(x) = x + f(x) o (x) = 2 x /2 i (x) = + f (x) ( x ) Therefore, we have that + f(x) = (x + 2 f(x)) /2 ( + f (x)). For the right-hand side, we must use the sum and product rules first, giving: d ( + x 2 f(x) 2) = 2xf(x) 2 + x 2 (f(x) 2 )

6 6 Again, to compute (f(x) 2 ), we must use the chain rule: o(x) = x 2 i(x) = f(x) o (x) = 2x i (x) = f (x) Therefore, (f(x) 2 ) = 2f(x)f (x) by the chain rule. Therefore, we see that the derivative of the equation at the top of the problem is: + f (x) 2 x + f(x) = 2xf(x)2 + 2x 2 f(x)f (x) Now we need to solve for f (x), and we will be finished. We have: + f (x) 2 x + f(x) = 2xf(x)2 + 2x 2 f(x)f (x) + f (x) = ( 2xf(x) 2 + 2x 2 f(x)f (x) ) ( 2 ) x + f(x) ( f (x) 2x 2 f(x)f (x) 2 ) ( x + f(x) = 2xf(x) 2 2 ) x + f(x) 2xf(x) (2 ) 2 x + f(x) f (x) = ( 2x 2 f(x) 2 ) x + f(x) Swapping dy cleaning up: for f (x) and y for f(x) gives the result, after a little dy = 4xy2 x + y 4x 2 y x + y 9. Find the equation of the line tangent to the curve x 2 +xy+y 2 = 3 at the point (, ). Solution. In order to find the equation of the tangent line at (, ), we need the slope of the line and a point on the line. Of course, we know immediately that (, ) is a point on the line. Recall that the slope of the tangent line is given by dy evaluated at (x, y) = (, ).

7 Differentiating, we have d ( x 2 + xy + y 2 = 3 ) = 2x + (y + xy ) + 2yy = 0 xy + 2yy = 2x y y (x + 2y) = 2x y y = 2x y x + 2y (If you don t see how the differentiation worked, write it out for yourself, using the worked-out examples above as a guide.) Evaluating at (x, y) = (, ), we have that the slope of the tangent line is 2 =. +2 Therefore, We have that the tangent line is Or, in slope-intercept form: This concludes the result. y = (x ) y = 2 x 7