Math 75 -- Fall 2002 -- Quiz 2 -- Warren D. Smith, ANSWERS ========================================================== Find a formula for the derivative of the following functions of x: 1. cos(ln( [x+2][x+5] )). ANSWER: use the chain rule twice and note that the derivative of 2 [x+2][x+5]=x + 7x + 10 is 2x+7. The final answer is sin(ln([x+2][x+5])) - ------------------- (2x+7). (x+2) (x+5) THAT AGAIN IN SLO-MO: The first time you use the chain rule, the outer function is cos and the inner function is ln( [x+2][x+5] ). The result is y'(x) = -sin(ln( [x+2][x+5] )) ln([x+2][x+5])'. Now we have to do the ' at the end. That is, we have to find out what ln([x+2][x+5])' is. That calls for a SECOND use of the chain rule, with outer=ln and inner(x)=[x+2][x+5]. The result is ln([x+2][x+5])' = (2x+7) / ([x+2][x+5]). Note 2x+7 is the derivative of (x+2)(x+5). SANITY CHECK at x=0: Answer says, y'(0) = -.520786. Slope near 0 is about [y(0.001)-y(0)]/.001 = -.520515. Sane! 2. 3 cos( --- - ln(2x+7) ) x 3 3 2 ANSWER (use chain rule): sin( --- - ln(2x+7) ) ( --- + ---- ) (and note that - signs cancel) x 2 2x+7 x SANITY CHECK at x=1: Answer says, y'(1) = 2.317702. Slope near 1 is about [y(1.001)-y(1)]/.001 = 2.3119256. Sane! ln(x) 3. ----- x-1 ANSWER (use quotient rule): (x-1)/x - ln(x) --------------- 2 (x-1) SANITY CHECK at x=2: Answer says, y'(2) = -.193147 Slope near 2 is about [y(2.001)-y(2)]/.001 = -.193147. Sane! 2 4. exp ( --- ) x 3+x ANSWER (use product and chain rules): 2 2 -2 exp(---) + x exp(---) ------ 3+x 3+x 2 (3+x) SANITY CHECK at x=0: Answer says, y'(0) = 1.94773. Slope near 0 is about [y(0.001)-y(0)]/.001 = 1.94730. 5. log ( 3 - x ) 5 ANSWER: use fact log (z) = ln(z) / ln(5). 5 Hence using rules for multiplying a function by a constant (1/ln(5) is the constant) and for differentiating ln, and use the chain rule (inner function maps x to 3-x, outer function is ln), we get: -1 1 ----------- = ----------- (3-x) ln(5) (x-3) ln(5) The -1 comes from the derivative of 3-x, which is -1, in the chain rule. The 1/(3-x) comes from ln'(z) = 1/z. The 1/ln(5) is just a constant scaling factor and differentiation has no effect on it. SANITY CHECK: As x gets more NEGATIVE, 3-x and hence ln(3-x) both increase, so this is a DECREASING function of x and the derivative should be NEGATIVE if x is large and negative. Yup. Find the values of the following limits: cos(x) - 1 6. lim ---------- x --> 0 2 x ANSWER: you get 0/0 if you just substitute x=0: useless. But we can employ L'Hopital to make this be lim -sin(x)/(2x). But that still gives 0/0 if just substitute x=0: still useless. OK employ L'Hopital AGAIN to make this be lim -cos(x)/2. Aha! Now we get by substituting x=0, just -1/2. (Sanity check: [cos(.01)-1]/.0001 = -.4999.) ln(x/2) 7. lim ------- x-->2 x-2 ANSWER: you get 0/0 if you just substitute x=2: useless. So we use L'Hopital to make this be lim (1/x)/1 (since ln(x/2)' = (2/x)/2 = 1/x by chain rule with inner x-->x/2 and outer=ln; also this can be seen by using ln(x/2) = ln(x)-ln(2), whose derivative is 1/x - 0 = 1/x. Also note, as far as the bottom is concerned, (x-2)' = 1). NOW we get by substituting x=2, just 1/2. [Sanity check: ln(2.01/2) /.01 = .499.] 8. Same problem, but x-->infinity instead of x-->2. ANSWER: 0 since the ln grows far more slowly than x. Hugeness rules: as x-->infinity, exponentials like x k A (any fixed A>1) grow far faster than x (any k). A B x grows far faster than x if A>B. k log(X) (to any base) grows much slower than x (any k>0). Also: can answer this problem by using L'Hopital: lim (1/x)/1 which clearly approaches 0 as x approaches infinity. [SANITY check: ln(1000/2) / (1000-2) = 0.006.] For the following rational function 3+4x y = -------- 2 1 + x find: 9. the (x,y) coordinates of all the roots (x at which y=0): ANSWER: y=0 when the top is 0 (when x = -3/4) [or bottom is infinity, which does not happen], so: (-3/4, 0) 10. ... and local maxes and local mins, By the quotient rule 2 2 2 [ 1 + x ] 4 - (3+4x) 2x - 4 x - 6x + 4 -2x -3x + 2 y' = ------------------------- = -------------- = 2 ----------- 2 2 2 2 2 2 [ 1 + x ] [ 1 + x ] [ 1 + x ] the maxes and mins can only occur at a critical point (where y'=0 or y' is undefined) i.e. where the top is 0 [or the bottom is 0 or infinity, neither of which happens]. So the top is 0, by the quadratic formula, when x = -(3 +or- 5)/4, i.e. when x = 1/2 or x = -2. MAX MIN y = 4 y = -1 SANITY CHECKS: x = .51 and x = .49 give y=3.999; x = -2.01 and x = -1.99 give y=-.99998. 11. ... and vertical asymptotes (x at which |y| --> infinity) ANSWER: there are no vertical asymptotes since the bottom of the fraction defining y, is never 0. 12. ... and horizontal asymptotes (y at which |x| --> infinity) ANSWER: when x --> +infinity, y --> 0. when x --> -infinity, y --> 0 also. 13. Find the x-coordinates of the inflection points of 4 3 2 y = x + 7 x - 3 x + 5 x + 1. For which x is this function concave-U and for which x is it concave-down? 3 2 y' = 4 x + 21 x - 6 x + 5. 2 2 y'' = 12 x + 42 x - 6 = 6 ( 2 x + 7 x - 1 ). So the inflection points are where y''=0, which is where (by the quadratic formula with A=2, B=7, C=-1; do not forget that - sign) -7 +or- sqrt(57) x = ---------------- since 57 = 49 + 8 4 It's concave-down if x lies BETWEEN these two roots, concave-U if x lies OUTSIDE of them, since y''>0 in the latter case. PROBLEM 14. What is the formula for the linear approximation to pi x y = cos(----) + 3 ln(x) 2 in the neighborhood of x=1? ANSWER: Since y' = 3/x - sin(pi x/2) (pi/2), y'(1) = 3-pi/2 is the slope at x=1. The value at x=1 is y(1) = 0+3ln(1)=0. So: y = (3-pi/2) (x-1). linear approx This is from y = y(B) + (x-B) y'(B) = y(1) + (x-1) y'(1) linear approx where B=basepoint since b=1 in this problem and then using the formulas for y(x) at x=1, and y'(x) at x=1 above. The fact that y(1)=0 here makes it especially easy. SANITY CHECK: x=1 gives y=0=yLinApprox. Good. x=1.1 gives y=.129 & yLinApprox=.143, which is only .014 off. PROBLEM 15. The derivative crosses 0 where the given function is flat (at its mins and maxes) and is above 0 when the given function is INCREASING... ------------------------end.