3 Rules for Fining Derivatives It is teious to compute a limit every time we nee to know the erivative of a function. Fortunately, we can evelop a small collection of examples an rules that allow us to compute the erivative of almost any function we are likely to encounter. Many functions involve quantities raise to a constant power, such as polynomials an more complicate combinations like y = (sinx) 4. So we start by examining powers of a single variable; this gives us a builing block for more complicate examples. º½ Ì ÈÓÛ Ö ÊÙÐ We start with the erivative of a power function, f(x) = x n. Here n is a number of any kin: integer, rational, positive, negative, even irrational, as in x π. We have alreay compute some simple examples, so the formula shoul not be a complete surprise: x xn = nx n 1. It is not easy to show this is true for any n. We will o some of the easier cases now, an iscuss the rest later. The easiest, an most common, is the case that n is a positive integer. To compute the erivative we nee to compute the following limit: x xn (x+) n x n. For a specific, fairly small value of n, we coul o this by straightforwar algebra. 55
56 Chapter 3 Rules for Fining Derivatives EXAMPLE 3.1.1 Fin the erivative of f(x) = x 3. x x3 (x+) 3 x 3. x 3 +3x 2 +3x 2 + 3 x 3. 3x 2 +3x 2 + 3. 3x2 +3x+ 2 = 3x 2. The general case is really not much harer as long as we on t try to o too much. The key is unerstaning what happens when (x+) n is multiplie out: (x+) n = x n +nx n 1 +a 2 x n 2 2 + ++a n 1 x n 1 + n. We know that multiplying out will give a large number of terms all of the form x i j, an in fact that i+j = n in every term. One way to see this is to unerstan that one metho for multiplying out (x+) n is the following: In every (x+) factor, pick either the x or the, then multiply the n choices together; o this in all possible ways. For example, for (x+) 3, there are eight possible ways to o this: (x+)(x+)(x+) = xxx+xx+xx+x +xx+x+x+ = x 3 +x 2 +x 2 +x 2 +x 2 +x 2 +x 2 + 3 = x 3 +3x 2 +3x 2 + 3 No matter what n is, there are n ways to pick in one factor an x in the remaining n 1 factors; this means one term is nx n 1. The other coefficients are somewhat harer to unerstan, but we on t really nee them, so in the formula above they have simply been calle a 2, a 3, an so on. We know that every one of these terms contains to at least the power 2. Now let s look at the limit: (x+) n x n x xn x n +nx n 1 +a 2 x n 2 2 + +a n 1 x n 1 + n x n nx n 1 +a 2 x n 2 2 + +a n 1 x n 1 + n nxn 1 +a 2 x n 2 + +a n 1 x n 2 + n 1 = nx n 1.
3.1 The Power Rule 57 Now without much trouble we can verify the formula for negative integers. First let s look at an example: EXAMPLE 3.1.2 Fin the erivative of y = x 3. Using the formula, y = 3x 3 1 = 3x 4. Here is the general computation. Suppose n is a negative integer; the algebra is easier to follow if we use n = m in the computation, where m is a positive integer. x xn = (x+) m x m x x m 1 (x+) 1 m x m x m (x+) m (x+) m x m x m (x m +mx m 1 +a 2 x m 2 2 + +a m 1 x m 1 + m ) (x+) m x m mx m 1 a 2 x m 2 a m 1 x m 2 m 1 ) (x+) m x m = mxm 1 x m x m = mxm 1 x 2m = mx m 1 2m = nx m 1 = nx n 1. We will later see why the other cases of the power rule work, but from now on we will use the power rule whenever n is any real number. Let s note here a simple case in which the power rule applies, or almost applies, but is not really neee. Suppose that f(x) = 1; remember that this 1 is a function, not merely a number, an that f(x) = 1 has a graph that is a horizontal line, with slope zero everywhere. So we know that f (x) = 0. We might also write f(x) = x 0, though there is some question about just what this means at x = 0. If we apply the power rule, we get f (x) = 0x 1 = 0/x = 0, again noting that there is a problem at x = 0. So the power rule works in this case, but it s really best to just remember that the erivative of any constant function is zero. Exercises 3.1. Fin the erivatives of the given functions. 1. x 100 2. x 100 1 3. x 5 4. xπ 5. x 3/4 6. x 9/7
58 Chapter 3 Rules for Fining Derivatives º¾ Ä Ò Ö ØÝ Ó Ø Ö Ú Ø Ú An operation is linear if it behaves nicely with respect to multiplication by a constant an aition. The name comes from the equation of a line through the origin, f(x) = mx, an the following two properties of this equation. First, f(cx) = m(cx) = c(mx) = cf(x), so the constant c can be move outsie or move through the function f. Secon, f(x+y) = m(x +y) = mx +my = f(x)+f(y), so the aition symbol likewise can be move through the function. The corresponing properties for the erivative are: an (cf(x)) = x cf(x) = c x f(x) = cf (x), (f(x)+g(x)) = x (f(x)+g(x)) = x f(x)+ x g(x) = f (x)+g (x). It is easy to see, or at least to believe, that these are true by thinking of the istance/spee interpretation of erivatives. If one object is at position f(t) at time t, we know its spee is given by f (t). Suppose another object is at position 5f(t) at time t, namely, that it is always 5 times as far along the route as the first object. Then it must be going 5 times as fast at all times. The secon rule is somewhat more complicate, but here is one way to picture it. Suppose a flat be railroa car is at position f(t) at time t, so the car is traveling at a spee of f (t) (to be specific, let s say that f(t) gives the position on the track of the rear en of the car). Suppose that an ant is crawling from the back of the car to the front so that its position on the car is g(t) an its spee relative to the car is g (t). Then in reality, at time t, the ant is at position f(t)+g(t) along the track, an its spee is obviously f (t)+g (t). We on t want to rely on some more-or-less obvious physical interpretation to etermine what is true mathematically, so let s see how to verify these rules by computation.
3.2 Linearity of the Derivative 59 We ll o one an leave the other for the exercises. f(x+)+g(x+) (f(x)+g(x)) (f(x)+g(x)) x f(x+)+g(x+) f(x) g(x) f(x+) f(x)+g(x+) g(x) ( f(x+) f(x) + g(x+) g(x) f(x+) f(x) g(x+) g(x) + lim = f (x)+g (x) This is sometimes calle the sum rule for erivatives. EXAMPLE 3.2.1 Fin the erivative of f(x) = x 5 +5x 2. We have to invoke linearity twice here: f (x) = x (x5 +5x 2 ) = x x5 + x (5x2 ) = 5x 4 +5 x (x2 ) = 5x 4 +5 2x 1 = 5x 4 +10x. ) Because it is so easy with a little practice, we can usually combine all uses of linearity into a single step. The following example shows an acceptably etaile computation. EXAMPLE 3.2.2 Fin the erivative of f(x) = 3/x 4 2x 2 +6x 7. f (x) = ( ) 3 x x 4 2x2 +6x 7 = x (3x 4 2x 2 +6x 7) = 12x 5 4x+6. Exercises 3.2. Fin the erivatives of the functions in 1 6. 1. 5x 3 +12x 2 15 2. 4x 5 +3x 2 5/x 2 3. 5( 3x 2 +5x+1) 4. f(x)+g(x), where f(x) = x 2 3x+2 an g(x) = 2x 3 5x 5. (x+1)(x 2 +2x 3) 6. 2 +3x 3 +12 (See section 2.1.) 7. Fin an equation for the tangent line to f(x) = x 3 /4 1/x at x = 2.
60 Chapter 3 Rules for Fining Derivatives 8. Fin an equation for the tangent line to f(x) = 3x 2 π 3 at x = 4. 9. Suppose the position of an object at time t is given by f(t) = 49t 2 /10 +5t +10. Fin a function giving the spee of the object at time t. The acceleration of an object is the rate at which its spee is changing, which means it is given by the erivative of the spee function. Fin the acceleration of the object at time t. 10. Let f(x) = x 3 an c = 3. Sketch the graphs of f, cf, f, an (cf) on the same iagram. n 11. The general polynomial P of egree n in the variable x has the form P(x) = a k x k = a 0 +a 1 x+...+a n x n. What is the erivative (with respect to x) of P? 12. Fin a cubic polynomial whose graph has horizontal tangents at ( 2,5) an (2,3). 13. Prove that x (cf(x)) = cf (x) using the efinition of the erivative. 14. Suppose that f an g are ifferentiable at x. Show that f g is ifferentiable at x using the two linearity properties from this section. k=0 º Ì ÈÖÓ ÙØ ÊÙÐ Consier the prouct of two simple functions, say f(x) = (x 2 +1)(x 3 3x). An obvious guess for the erivative of f is the prouct of the erivatives of the constituent functions: (2x)(3x 2 3) = 6x 3 6x. Is thiscorrect? We can easily check, by rewriting f an oing the calculationinaway that isknownto work. First, f(x) = x 5 3x 3 +x 3 3x = x 5 2x 3 3x, an then f (x) = 5x 4 6x 2 3. Not even close! What went wrong? Well, nothing really, except the guess was wrong. So the erivative of f(x)g(x) is NOT as simple as f (x)g (x). Surely there is some rule for such a situation? There is, an it is instructive to iscover it by trying to o the general calculation even without knowing the answer in avance. f(x+)g(x+) f(x)g(x) (f(x)g(x)) x f(x+)g(x+) f(x+)g(x)+f(x+)g(x) f(x)g(x) f(x+)g(x+) f(x+)g(x) f(x+)g(x) f(x)g(x) + lim f(x+)g(x+) g(x) + lim = f(x)g (x)+f (x)g(x) f(x+) f(x) g(x) A couple of items here nee iscussion. First, we use a stanar trick, a an subtract the same thing, to transform what we ha into a more useful form. After some rewriting, we realize that we have two limits that prouce f (x) an g (x). Of course, f (x) an
3.3 The Prouct Rule 61 g (x) must actually exist for this to make sense. We also replace lim f(x+) with f(x) why is this justifie? What we really nee to know here is that lim f(x+) = f(x), or in the language of section 2.5, that f is continuous at x. We alreay know that f (x) exists (or the whole approach, writing the erivative of fg in terms of f an g, oesn t make sense). This turns out to imply that f is continuous as well. Here s why: lim f(x+) (f(x+) f(x)+f(x)) f(x+) f(x) + lim f(x) = f (x) 0+f(x) = f(x) To summarize: the prouct rule says that x (f(x)g(x)) = f(x)g (x)+f (x)g(x). Returning to the example we starte with, let f(x) = (x 2 +1)(x 3 3x). Then f (x) = (x 2 + 1)(3x 2 3) + (2x)(x 3 3x) = 3x 4 3x 2 + 3x 2 3 + 2x 4 6x 2 = 5x 4 6x 2 3, as before. In this case it is probably simpler to multiply f(x) out first, then compute the erivative; here s an example for which we really nee the prouct rule. EXAMPLE 3.3.1 Compute the erivative of f(x) = x 2 2. We have alreay compute x 2 = x. Now 2 f (x) = x 2 x +2x 2 = x3 +2x( 2 ) 2 2 = 3x3 +1250x. 2 Exercises 3.3. In 1 4, fin the erivatives of the functions using the prouct rule. 1. x 3 (x 3 5x+10) 2. (x 2 +5x 3)(x 5 6x 3 +3x 2 7x+1) 3. x 2 2 4. x 20 5. Use the prouct rule to compute the erivative of f(x) = (2x 3) 2. Sketch the function. Fin an equation of the tangent line to the curve at x = 2. Sketch the tangent line at x = 2.
62 Chapter 3 Rules for Fining Derivatives 6. Suppose that f, g, an h are ifferentiable functions. Show that (fgh) (x) = f (x)g(x)h(x)+ f(x)g (x)h(x)+f(x)g(x)h (x). 7. State an prove a rule to compute (fghi) (x), similar to the rule in the previous problem. Prouct notation. Suppose f 1,f 2,...f n are functions. The prouct of all these functions can be written n f k. k=1 This is similar to the use of to enote a sum. For example, an 5 f k = f 1 f 2 f 3 f 4 f 5 k=1 n k = 1 2... n = n!. k=1 We sometimes use somewhat more complicate conitions; for example n k=1,k j enotes the prouct of f 1 through f n except for f j. For example, 5 x k = x x 2 x 3 x 5 = x 11. k=1,k 4 8. The generalize prouct rule says that if f 1,f 2,...,f n are ifferentiable functions at x then n n n f k (x) = f j(x) x f k (x). k=1 j=1 f k k=1,k j Verify that this is the same as your answer to the previous problem when n = 4, an write out what this says when n = 5. º Ì ÉÙÓØ ÒØ ÊÙÐ What is the erivative of (x 2 +1)/(x 3 3x)? More generally, we like to have a formula to compute the erivative of f(x)/g(x) if we alreay know f (x) an g (x). Instea of attacking this problem hea-on, let s notice that we ve alreay one part of the problem: f(x)/g(x) = f(x) (1/g(x)), that is, this is really a prouct, an we can compute the erivative if we know f (x) an (1/g(x)). So really the only new bit of information we nee is (1/g(x)) in terms of g (x). As with the prouct rule, let s set this up an see how
far we can get: x 1 g(x) 1 g(x+) 1 g(x) g(x) g(x+) g(x+)g(x) g(x) g(x+) g(x + )g(x) g(x+) g(x) 1 g(x + )g(x) = g (x) g(x) 2 Now we can put this together with the prouct rule: f(x) x g(x) = (x) 1 f(x) g g(x) 2 +f (x) g(x) = f(x)g (x)+f (x)g(x) g(x) 2 3.4 The Quotient Rule 63 = f (x)g(x) f(x)g (x) g(x) 2. EXAMPLE 3.4.1 Compute the erivative of (x 2 +1)/(x 3 3x). x 2 +1 xx 3 3x = 2x(x3 3x) (x 2 +1)(3x 2 3) (x 3 3x) 2 = x4 6x 2 +3 (x 3 3x) 2. It is often possible to calculate erivatives in more than one way, as we have alreay seen. Since every quotient can be written as a prouct, it is always possible to use the prouct rule to compute the erivative, though it is not always simpler. EXAMPLE 3.4.2 Fintheerivativeof 2 / xintwo ways: using thequotient rule, an using the prouct rule. Quotient rule: x 2 x = x( x/ 2 ) 2 1/(2 x). x Note that we have use x = x 1/2 to compute the erivative of x by the power rule. Prouct rule: 2 x 1/2 = x 2 1 x 2 x 3/2 + 2 x 1/2. With a bit of algebra, both of these simplify to x 2 +625 2 2 x 3/2.
64 Chapter 3 Rules for Fining Derivatives Occasionally you will nee to compute the erivative of a quotient with a constant numerator, like 10/x 2. Of course you can use the quotient rule, but it is usually not the easiest metho. If we o use it here, we get 10 xx 2 = x2 0 10 2x x 4 = 20 x 3, since the erivative of 10 is 0. But it is simpler to o this: 10 xx = 2 x 10x 2 = 20x 3. Amittely, x 2 is a particularly simple enominator, but we will see that a similar calculation is usually possible. Another approach is to remember that 1 xg(x) = g (x) g(x) 2, but this requires extra memorization. Using this formula, 10 xx = 10 2x 2 x. 4 Note that we first use linearity of the erivative to pull the 10 out in front. Exercises 3.4. Fin the erivatives of the functions in 1 4 using the quotient rule. 1. 3. x 3 x 3 5x+10 2. x 2 +5x 3 x 5 6x 3 +3x 2 7x+1 x 4. 2 2 x 20 5. Fin an equation for the tangent line to f(x) = (x 2 4)/(5 x) at x = 3. 6. Fin an equation for the tangent line to f(x) = (x 2)/(x 3 +4x 1) at x = 1. 7. Let P be a polynomial of egree n an let Q be a polynomial of egree m (with Q not the zero polynomial). Using sigma notation we can write n m P = a k x k, Q = b k x k. k=0 Use sigma notation to write the erivative of the rational function P/Q. 8. The curve y = 1/(1+x 2 ) is an example of a class of curves each of which is calle a witch of Agnesi. Sketch the curve an fin the tangent line to the curve at x = 5. (The wor k=0
3.5 The Chain Rule 65 witch here is a mistranslation of the original Italian, as escribe at an http://mathworl.wolfram.com/witchofagnesi.html http://instructional1.calstatela.eu/sgray/agnesi/ WitchHistory/Historynamewitch.html.) 9. If f (4) = 5, g (4) = 12, (fg)(4) = f(4)g(4) = 2, an g(4) = 6, compute f(4) an f x g at 4. º Ì Ò ÊÙÐ So far we have seen how to compute the erivative of a function built up from other functions by aition, subtraction, multiplication an ivision. There is another very important way that we combine simple functions to make more complicate functions: function composition, as iscusse in section 2.3. For example, consier 2. This function has many simpler components, like 625 an x 2, an then there is that square root symbol, so the square root function x = x 1/2 is involve. The obvious question is: can we compute the erivative using the erivatives of the constituents 2 an x? We can inee. In general, if f(x) an g(x) are functions, we can compute the erivatives of f(g(x)) an g(f(x)) in terms of f (x) an g (x). EXAMPLE 3.5.1 Form the two possible compositions of f(x) = x an g(x) = 625 x 2 an compute the erivatives. First, f(g(x)) = 2, an the erivative is x/ 2 as we have seen. Secon, g(f(x)) = 625 ( x) 2 = with erivative 1. Of course, these calculations o not use anything new, an in particular the erivative of f(g(x)) was somewhat teious to compute from the efinition. Suppose we want the erivative of f(g(x)). Again, let s set up the erivative an play some algebraic tricks: f(g(x+)) f(g(x)) f(g(x)) x f(g(x+)) f(g(x)) g(x+)) g(x) g(x+)) g(x) Now we see immeiately that the secon fraction turns into g (x) when we take the limit. The first fraction is more complicate, but it too looks something like a erivative. The enominator, g(x + )) g(x), is a change in the value of g, so let s abbreviate it as
66 Chapter 3 Rules for Fining Derivatives g = g(x+)) g(x), which also means g(x+) = g(x)+ g. This gives us lim f(g(x)+ g) f(g(x)). g As goes to 0, it is also true that g goes to 0, because g(x+) goes to g(x). So we can rewrite this limit as f(g(x)+ g) f(g(x)) lim. g 0 g Now this looks exactly like a erivative, namely f (g(x)), that is, the function f (x) with x replace by g(x). If this all withstans scrutiny, we then get x f(g(x)) = f (g(x))g (x). Unfortunately, thereisasmallflaw intheargument. Recallthatwhatwemeanbylim involves what happens when is close to 0 but not equal to 0. The qualification is very important, since we must be able to ivie by. But when is close to 0 but not equal to 0, g = g(x+)) g(x) is close to 0 an possibly equal to 0. This means it oesn t really make sense to ivie by g. Fortunately, it is possible to recast the argument to avoi this ifficulty, but it is a bit tricky; we will not inclue the etails, which can be foun in many calculus books. Note that many functions g o have the property that g(x + ) g(x) 0 when is small, an for these functions the argument above is fine. The chain rule has a particularly simple expression if we use the Leibniz notation for the erivative. The quantity f (g(x)) is the erivative of f with x replace by g; this can be written f/g. As usual, g (x) = g/x. Then the chain rule becomes f x = f g g x. This looks like trivial arithmetic, but it is not: g/x is not a fraction, that is, not literal ivision, but a single symbol that means g (x). Nevertheless, it turns out that what looks like trivial arithmetic, an is therefore easy to remember, is really true. It will take a bit of practice to make the use of the chain rule come naturally it is more complicate than the earlier ifferentiation rules we have seen. EXAMPLE 3.5.2 Compute the erivative of 2. We alreay know that the answer is x/ 2, compute irectly from the limit. In the context of the chain rule, we have f(x) = x, g(x) = 625 x 2. We know that f (x) = (1/2)x 1/2, so
3.5 The Chain Rule 67 f (g(x)) = (1/2)( 2 ) 1/2. Note that this is a two step computation: first compute f (x), then replace x by g(x). Since g (x) = 2x we have f (g(x))g (x) = 1 2 2( 2x) = x. 2 EXAMPLE 3.5.3 Compute the erivative of 1/ 2. This is a quotient with a constant numerator, so we coul use the quotient rule, but it is simpler to use the chain rule. The function is ( 2 ) 1/2, the composition of f(x) = x 1/2 an g(x) = 2. We compute f (x) = ( 1/2)x 3/2 using the power rule, an then f (g(x))g (x) = 1 2( 2 ) 3/2( 2x) = x ( 2 ) 3/2. Inpractice, ofcourse, youwillneetousemorethanoneoftheruleswehaveevelope to compute the erivative of a complicate function. EXAMPLE 3.5.4 Compute the erivative of f(x) = x2 1 x x 2 +1. The last operation here is ivision, so to get starte we nee to use the quotient rule first. This gives f (x) = (x2 1) x x 2 +1 (x 2 1)(x x 2 +1) x 2 (x 2 +1) = 2x2 x 2 +1 (x 2 1)(x x 2 +1) x 2 (x 2. +1) Now we nee to compute the erivative of x x 2 +1. This is a prouct, so we use the prouct rule: x x x 2 +1 = x x2 +1+ x x 2 +1. Finally, we use the chain rule: x2 +1 = x x (x2 +1) 1/2 = 1 2 (x2 +1) 1/2 (2x) = x x2 +1.
68 Chapter 3 Rules for Fining Derivatives An putting it all together: f (x) = 2x2 x 2 +1 (x 2 1)(x x 2 +1) x 2 (x 2. +1) ( ) 2x 2 x 2 +1 (x 2 x 1) x x2 +1 + x 2 +1 = x 2 (x 2. +1) This can be simplifie of course, but we have one all the calculus, so that only algebra is left. EXAMPLE 3.5.5 Compute the erivative of 1+ 1+ x. Here we have a more complicate chain of compositions, so we use the chain rule twice. At the outermost layer we have the function g(x) = 1+ 1+ x plugge into f(x) = x, so applying the chain rule once gives 1+ 1+ x = 1 ( 1+ 1+ ) 1/2 ( x 1+ 1+ ) x. x 2 x Now we nee the erivative of So the original erivative is 1+ x. Using the chain rule again: 1+ x = 1 ( ) 1/2 1 1+ x x 2 2 x 1/2. 1+ 1+ x = 1 ( 1+ 1+ ) 1/2 1 ( ) 1/2 1 x 1+ x x 2 2 2 x 1/2. = 1 8 x 1+ x 1+ 1+ x Using the chain rule, the power rule, an the prouct rule, it is possible to avoi using the quotient rule entirely.
3.5 The Chain Rule 69 EXAMPLE 3.5.6 Computetheerivativeoff(x) = x3 x 2 +1. Writef(x) = x3 (x 2 +1) 1, then f (x) = x 3 x (x2 +1) 1 +3x 2 (x 2 +1) 1 = x 3 ( 1)(x 2 +1) 2 (2x)+3x 2 (x 2 +1) 1 = 2x 4 (x 2 +1) 2 +3x 2 (x 2 +1) 1 = 2x4 (x 2 +1) 2 + 3x2 x 2 +1 = 2x4 (x 2 +1) + 3x2 (x 2 +1) 2 (x 2 +1) 2 = 2x4 +3x 4 +3x 2 (x 2 +1) 2 = x4 +3x 2 (x 2 +1) 2 Note that we alreay ha the erivative on the secon line; all the rest is simplification. It is easier to get to this answer by using the quotient rule, so there s a trae off: more work for fewer memorize formulas. Exercises 3.5. Fin the erivatives of the functions. For extra practice, an to check your answers, o some of these in more than one way if possible. 1. x 4 3x 3 +(1/2)x 2 +7x π 2. x 3 2x 2 +4 x 3. (x 2 +1) 3 4. x 169 x 2 5. (x 2 4x+5) 25 x 2 6. r2 x 2, r is a constant 1 7. 1+x4 8.. 5 x 9. (1+3x) 2 (x 2 +x+1) 10. (1 x) 25 x 2 169 11. 12. x x x 13. x3 x 2 (1/x) 14. 100/(100 x 2 ) 3/2 3 15. x+x3 16. (x 2 +1) 2 + 1+(x 2 +1) 2 17. (x+8) 5 18. (4 x) 3 19. (x 2 +5) 3 20. (6 2x 2 ) 3 21. (1 4x 3 ) 2 22. 5(x+1 1/x) 23. 4(2x 2 x+3) 2 1 24. 1+1/x 3 25. 4x 2 2x+1 26. (x2 +1)(5 2x)/2
70 Chapter 3 Rules for Fining Derivatives 27. (3x 2 +1)(2x 4) 3 28. 29. 31. 33. x 2 1 x 2 +1 2x 1 x 2 3x 1 4x 2 1 (2x+1)(x 3) 35. (2x+1) 3 (x 2 +1) 2 x+1 x 1 30. (x 1)(x 2) x 3 32. 3(x2 +1)(2x 2 1)(2x+3) 34. ((2x+1) 1 +3) 1 36. Fin an equation for the tangent line to f(x) = (x 2) 1/3 /(x 3 +4x 1) 2 at x = 1. 37. Fin an equation for the tangent line to y = 9x 2 at (3,1). 38. Fin an equation for the tangent line to (x 2 4x+5) 25 x 2 at (3,8). 39. Fin an equation for the tangent line to (x2 +x+1) at (2, 7). (1 x) 40. Fin an equation for the tangent line to (x 2 +1) 2 + 1+(x 2 +1) 2 at (1, 4+ 5).