x 2. By rewriting f as f (x) = 4x 1 and using the Product Rule, x 2 = 4 f = 2e x csc x csc x csc x = 2e x ( csc x cot x) + 2e x csc x + cot x) sin 1 x

Transcription

1 MTH 07 Final Eam Study Questions: Derivatives, KEY 1. Find the derivative of f = 4 using the Quotient Rule. Fin using the Proct Rule. Fin without using either rule. Solution: Using the Quotient rule, f = = = 0 4 = 4. By rewriting f as f = 4 1 and using the Proct Rule, f = = = 4 By still using f rewritten as f = 4 1 and using the Constant Multiple Rule, f = 4 1 = 4.. If f = e csc 6 + cot sin1, fin Solution: We can consider one term at a time: in other words, by using f = e csc g 6 + cot sin1 y we have f = g y, thus f = g y. The derivative of g is found by using the Proct Rule. First, here s how I d like to think of g as the proct of two functions, though this is not the only way to do this: g = e csc Then g = e csc + e csc = e csc cot + e csc Now how about y? Since y has s in BOTH the base AND eponent, the ONLY way to take the derivative is by logarithmic differentiation. y = 6 + cot sin1 ln y = ln 6 + cot sin1 ln y = sin 1 ln6 + cot Now differentiate both sides with respect to, keeping in mind that y is a function of on the left side: By rewriting the algebra, y = 1 sin1 6 + cot 1 csc + ln6 + cot 1 y Multiply by y on both sides: y y = csc sin cot csc y sin 1 = y cot ln6 + cot + 1 ln6 + cot 1 Replace y with 6 + cot sin1 to get y = 6 + cot sin1 csc sin cot ln6 + cot 1 Now, let s piece everything together: f = g y = e csc cot + e csc g 6 + cot sin1 csc sin cot + ln6 + cot 1 y - page 1-

2 MTH 07 Final Eam Study Questions: Derivatives, KEY 3. If y = tan 1 + 5sin 3 + log 4, find d y Solution: This is to remind you that in Leibnitz notation, the derivative of y with respect to input is written as d y. The same thing in Newton notation is y or y. See the top half of page 157. d y = log 4 5cos 5sin 3 ln log 4 1 ln4 4. If y = sec tan, find y Solution: y = sec 4 tan tan ln3 sec Be sure that you have the factor of tan 4 in the first term and the factwor of ln3 in the second term. It is easy to forget to include these in the Chain Rule. If you disagree that these belong, carefully write out f u = secu and g = 4, thus [f g ] = f g g, and write out all the steps systematically. 5. If = t sin t, find d dt Solution: Writing dt is the Leibnitz notation for the first derivative of with respect to t and writing d dt is the second derivative of with respect to t. Read the top half of page 161 in the book for a refresher. In Newton notation, dt would be written and d would be written. What s unusual about this dt problem but it s very common in later Calc classes is the use of as the output and t as the input, as usually is the input. By the Proct Rule, = t cos t + sin t dt To find the second derivative, we ll need the Proct Rule on the first term: d = t sin t + cos t dt + cos t = cos t t sin t d t cos t 6. If f =, what is f 100? Solution: This is Newton notation for the 100th derivative. Keeping in mind that ln is just a constant, f 100 = ln 100. Do not use the Proct Rule for this problem AT ALL! Use the constant multiple rule!!!!!!! Also, be sure to write f 100, not f 100, which means something else. - page -

3 MTH 07 Final Eam Study Questions: Derivatives, KEY log6 7. If f = tan sinm dm, find the derivative of f with respect to. Solution: Normally, for FTC1 problems, the integrand of the integral is calle. It would be helpful to first go through the self-guided study on FTC1 click and its key click before trying this and the net problem. Let g = log6 sinm dm. Then f = tang. We are asked to find. By the Chain Rule, =. The harder task here is to find. Rewrite g by splitting the integral: g = = = log6 07 log6 sinm dm sinm dm + sinm dm 07 h log6 07 sinm dm sinm dm 07 k I have intentionally NOT included the minus sign in the definition of k. Thus, g = h k. You might have defined things a little differently, but as long as you account for the minus sign correctly, great! By letting u = log 6, we can rewrite h as follows: u h = sinm dm 07 We were trying to find. Since g = h k, then = u, but to find, we ll just directly use FTC1. = = = sinu 1 ln6 Going back to the derivative of f with respect to, = sec g sinlog6 ln6. To find, we ll use the Chain Rule via sin sin = sinlog 6 sin. ln6 Now just replace the g, since there was no g in the original problem. We ve written g in several ways. Be sure to use a version of g which does NOT have a u in it because again, u was not in the original question. I ll use the shortest such epression I see three on this page. log6 sec = sinm sinlog6 dm sin ln6 - page 3-

4 MTH 07 Final Eam Study Questions: Derivatives, KEY Short version: Before moving on to the net one, here s a shorter version of the previous solution. I ll use all the same letters: Let u = log 6. Also, let g = log6 u sinm dm = sinm dm 07 h Since f = tang and g = h k, by the Chain Rule, sinm dm 07 k = = since g =hk = = sec g sinu 1 ln6 log6 = sec sinm dm sin sinlog6 ln6 sin - page 4-

5 MTH 07 Final Eam Study Questions: Derivatives, KEY 3 8. If q = sec sinr + r dr, find the derivative of q with respect to. ln Solution: Let g = as follows: 3 ln g = sinr + r dr, so that q = secg. By letting u = ln and v = 3, we can rewrite g v Since g = h k, we have = u sinr + r dr = v sinr + r dr 10 h, which we ll use below: dq = dq = dq sinv + v = secg tang dq which, after substituting away all the letters we introced, becomes: 3 3 sec sinr + r dr tan sinr + r dr ln ln 3 Be sure to have the factor of tan sinr + r dr in your final answer. ln u sinr + r dr 10 k 3 ln3 sinu + u 1 sin ln3 sinln + ln 1 9. The position of a buoy in a lake is given by st = 0 + sin πt 4 in feet, where t is in seconds. What is the velocity of the buoy at time t = 1 seconds? Be sure to state units. Solution: The velocity is the rate of change of position, so the question is asking for s 1. First, let s find s t. s t = cos πt 4 π 4 = π πt cos 4 so s 1 = π π 1 cos 4 = π cos3π = π. The velocity of the buoy is π ft/sec. Or, we can say that the buoy has a downward speed of π ft/sec. - page 5-

6 MTH 07 Final Eam Study Questions: Derivatives, KEY 10. Find the absolute maimum and absolute minimum points of f = + sin on the interval [0,10π]. Solution: Since f is continuous and the interval is closed, use the Closed Interval Method. Set this equal to zero to find any critical numbers: f = 1 + cos. 1 + cos = 0 cos = 1 The solutions to this for -values in the interval [0,10π] are = π, = 3π, = 5π, = 7π, and = 9π. Why does the Closed Inteval Method have us look at these -values? These are the parts of the graph which are level / flat. Consider the -values = π, = 3π, = 5π, = 7π, and = 9π and as well, consider the two endpoints = 0 and = 10π. By plugging in these numbers into the function f, we ll see which output is biggest/smallest. Be sure to plug in to f, not f. At this point, we are done with the derivative function. f 0 = 0 + sin0 = 0 f π = π + sinπ = π f 3π = 3π + sin3π = 3π f 5π = 5π + sin5π = 5π f 7π = 7π + sin7π = 7π f 9π = 9π + sin9π = 9π f 10π = 10π + sin10π = 10π Since the largest y-value is 10π, the point 10π,10π is the absolute maimum point on the interval [0,10π]. Since the largest y-value is 0π, the point 0,0 is the absolute maimum point. 11. Show that f = tan 1 4 has at most one root. Solution: Suppose that f has two or more roots. Call one of the roots a and one of them b. Then f a = 0 an b = 0. Since f is continuous on [a,b] and differentiable on a,b, Rolle s Theorem applies. Rolle s Theorem guarantees that there is an -value c in the interval a,b such that f c = 0. Since f = sec 1 4, we get f c = sec c 1 4. Thus, the statement that f c = 0 can be rewritten as sec c 1 4 = 0. Rewrite: sec c = 1 4 secc = ± 1 cosc = ± But there are no solutions to cosc = or cosc =. Therefore, the assumption that f has two or more roots is false. In other words, f has at most one root. 1. Show that f = tan 1 4 has at least one root in the interval [ π 3, π 3 ] Solution: The function f is continuous on the interval [ π 3, π 3 ]. Using a = π 3 and b = π 3, f a = f π 3 = tan π π 3 = 3 + π 1 f b = f π 3 = tan π π 3 = 3 π 1 Pick N = 0, which is in between 3 + π 1 and 3 π 1. By the Intermediate Value Theorem, there is an -value c in the interval a,b such that f c = 0. - page 6-

7 MTH 07 Final Eam Study Questions: Derivatives, KEY 13. The limit 1 + h 10 1 lim h 0 h represents the derivative of some function f at some number a. State such an f and a. Solution: f = 10 and a = Find d y if y is a function of according to the equation 3 y + y 3 = 5. Solution: Differentiate both sides implictly with respect to. 3 y + y 3 = 5 3 y + y + 3y y = 0 3 y y + 3y y = 0 y 3y = y 3 y = y 3 3y 15. Show that the curve y = e has no tangent line with slope. Solution: The slope of a tangent line is the derivative, so differentiate: y = e y = e To say that there is a slope of two for the tangent line means that there is an where y =. Be sure that you re not plugging in =. So, e = But the problem is that e and are both positive, so the left side is always bigger than 3. - page 7-