INVERSE FUNCTIONS DERIVATIVES Recall the steps for computing y implicitly: (1) Take of both sies, treating y like a function. (2) Expan, a, subtract to get the y terms on one sie an everything else on the other. (3) Factor out y an ivie both sies by its coefficient. Warmup: Use implicit ifferentiation to compute y for the following functions: 1. e y = xy 2. cos(y) = x + y Inverse functions. Let f(x) be a function. Recall f 1 (x) is the inverse function of f(x), meaning y = f 1 (x) if an only if x = f(y). You can also think of f 1 (x) as unoing f(x), i.e. For example, ln(x) is the inverse of e x, since f(f 1 (x)) = x an f 1 (f(x)) = x. y = ln(x) if an only if x = e y. So e ln(x) = x an ln(e x ) = x for x > 0. Problem 1: Using a calculator or computer, fill out the following table. x 0 1 2 10 1 2 10 e x ln(x) ln(e x ) e ln(x) 1
2 INVERSE FUNCTIONS DERIVATIVES What o you notice when x is negative? Explain the pattern. First: Rewrite y = ln x as e y = x. Next: Take a erivative of both sies of e y = x: ey = x =. Next: Set your answers equal to each other an solve for y. y = But there s still a problem! We aske what is the erivative of ln(x)? an got back an answer with y in it! To fix this problem, plug in y = e x an simplify. Finally, you shoul then get ln(x) = 1 x.
INVERSE FUNCTIONS DERIVATIVES 3 Problem 2: Compare the graphs of 1/x an ln(x) an check that it makes sense that ln(x) = 1 x. Problem 3: Using your new ientity, ln(x) = 1 x, compute the following. (1) ln x2 (2) ln(sin(x2 )) (3) log 3(x) [hint: log a x = ln x ln a ] (4) ln(f(x)), where f(x) is a ifferentiable function. Computing erivatives of inverses in general. In the case where y = ln(x), we use the fact that ln(x) = f 1 (x), where f(x) = e x, an got In general, calculating f 1 (x): First: Rewrite y = f 1 (x) as f(y) = x. ln(x) = 1 e ln(x). Next: Use implicit ifferentiation to compute y : f (y) y = 1 so y = 1 f (y). Finally: Plug back in y = f 1 (x) to get an answer in terms of just x: y = 1 f (f 1 (x)).
4 INVERSE FUNCTIONS DERIVATIVES Problem 4: Check that this ientity makes sense by applying it to the following examples. (1) Compute x by noting that if f(x) = x 2 then f 1 (x) = x. (2) Compute 3 x by noting that if f(x) = x 3 then f 1 (x) = 3 x. (3) Compute ln(x) by noting that if f(x) = e x then f 1 (x) = ln(x).
INVERSE FUNCTIONS DERIVATIVES 5 Inverse trig functions. Recall that we have two notations for inverse trig functions: f(x) sin(x) cos(x) tan(x) sec(x) csc(x) cot(x) f 1 (x) sin 1 (x) = arcsin(x) cos 1 (x) = arccos(x) tan 1 (x) = arctan(x) sec 1 (x) = arcsec(x) csc 1 (x) = arccsc(x) cot 1 (x) = arccot(x) Review the notes for the graphs of these functions. Also, recall the erivatives of our favorite trig functions: f(x) sin(x) cos(x) tan(x) sec(x) csc(x) cot(x) f (x) cos(x) sin(x) sec 2 (x) sec(x) tan(x) csc(x) cot(x) csc 2 (x) To compute the erivative of arcsin(x): First: Rewrite y = arcsin(x) as sin(y) = x. Next: Use implicit ifferentiation to compute y : cos(x) y = 1 so y = 1 cos(y). Finally: Plug back in y = arcsin(x) to get an answer in terms of just x: y = 1 cos(arcsin(x)). Problem 5: Use implicit ifferentiation to calculate the erivative of arctan(x). Problem 6: Use the rule f 1 1 (x) = f (f 1 (x)) to check the answers to arcsin(x) an arctan(x).
6 INVERSE FUNCTIONS DERIVATIVES Problem 7: Use the rule f 1 1 (x) = f (f 1 (x)) to calculate the erivatives of the other inverse trig functions: (1) arccos(x) (2) arcsec(x) (3) arccsc(x) (4) arccot(x) We ll go over how to simplify these in class on Tuesay. Solutions to warmup: (1) e y = xy: Take So of both sies to fin ey y = x y + y. y = e y y x y = y (ey x), (2) cos(y) = x + y: Take y as before: sin(y) = 1 + y. So implying y y (sin(y) + 1) = 1, an so = y = 1 sin(y) + 1. y e y x
INVERSE FUNCTIONS DERIVATIVES 7 Answers Answers to Problem 1: Using a calculator or computer, fill out the following table. x 0 1 2 10 1 2 10 e x 1 2.718... 7.389... 22026.318... 0.368... 0.135... 0.000045... ln(x) Unef 0 0.693... 2.303... Unef Unef Unef ln(e x ) 0 1 2 10 1 2 10 e ln(x) Unef 1 2 10 Unef Unef Unef When x is positive, we have ln(e x ) = x an e ln(x) = x. But when x is non-positive (0 or negative), then ln(e x ) = x, but e ln(x) is unefine. Answers to Problem 3: (1) ln x2 = 2x x 2 = 2 x (2) (3) (4) ln(sin(x2 )) = 2x cos(x2 ) sin(x 2 ) log 3(x) = 1 x ln(3) ln(f(x)) = f (x) f(x) Answers to Problem 4: (1) Compute x by noting that if f(x) = x 2 then f 1 (x) = x. Answer. If f 1 (x) = x, then f(x) = x 2, so that f (x) = 2x. Then 1 x = 2 x. Compare this to x = x1/2 = 1 2 x 1/2 = 1 2 x.
8 INVERSE FUNCTIONS DERIVATIVES (2) Compute 3 x by noting that if f(x) = x 3 then f 1 (x) = 3 x. Answer. If f 1 (x) = 3 x, then f(x) = x 3, so that f (x) = 3x 2. Then 3 1 x = 3( 3 x) 2 = 1 3x 2/3. Compare this to 3 x = x1/3 = 1 3 x 2/3 = 1 3x 2/3. (3) Compute ln(x) by noting that if f(x) = e x then f 1 (x) = ln(x). Answer. If f 1 (x) = ln(x), then f(x) = e x, so that f (x) = e x. Then ln(x) = 1 e ln(x) = 1 x. Problem 5: Use implicit ifferentiation to calculate the erivative of arctan(x). Answer. We have y = arctan(x) if an only if tan(y) = x. Taking So tan(y) = sec2 (y) y an x = 1. of both sies gives sec 2 (y) y = 1, so that y = 1 (sec(y)) 2 = 1 (sec(tan(x))) 2. Problem 6: Use the rule f 1 (x) = to check the answers to arcsin(x) an arctan(x). 1 f (f 1 (x)) Answer. If f 1 (x) = arcsin(x), then f(x) = sin(x), so that f (x) = cos(x). Then arcsin(x) = 1 cos(arcsin(x)). If f 1 (x) = arctan(x), then f(x) = tan(x), so that f (x) = sec 2 (x). Then arctan(x) = 1 sec 2 (arctan(x)).
INVERSE FUNCTIONS DERIVATIVES 9 Problem 7: Use the rule f 1 1 (x) = f (f 1 (x)) to calculate the erivatives of the other inverse trig functions: (1) arccos(x) Answer. If f 1 (x) = arccos(x), then f(x) = cos(x), so that f (x) = sin(x). Then (2) arcsec(x) arcsin(x) = 1 sin(arccos(x)). Answer. If f 1 (x) = arcsec(x), then f(x) = sec(x), so that f (x) = sec(x) tan(x). Then (3) arccsc(x) arcsec(x) = 1 sec(arcsec(x)) tan(arcsec(x)). Answer. If f 1 (x) = arccsc(x), then f(x) = csc(x), so thar f (x) = csc(x) cot(x). Then (4) arccot(x) arccsc(x) = 1 csc(arcsec(x)) cot(arcsec(x)). Answer. If f 1 (x) = arccot(x), then f(x) = cot(x), so that f (x) = csc 2 (x). Then arccot(x) = 1 csc 2 (arccot(x)).