Derivatives of Inverse Functions

Transcription

1 Derivatives of Inverse Functions Implicit differentiation enables us to determine the derivatives of inverse functions. determine the derivatives of arcsin, arccos, arctan, and ln. In this lecture, we To find the derivative of arcsin Let f) = sin, π π. Its inverse is f ) = arcsin, also written as sin ), which you should not mistake for / sin ). Its graph is shown below. arcsin Graph of arcsin To simplify notation, write f ) = arcsin as y = arcsin, which one may read as "y is the angle whose sine is "). Thus sin y = ) Differentiate both sides of ) implicitly and solve for the derivative of y. You should get d = cos y. But we need a derivative epressed in terms of. To get it, we turn to the identity cos y + sin y = It implies that cos y = ± sin y = ±. Since the slope of the tangent at any point on the graph of y is positive, see its graph), we must take the positive sign. Therefore if y = arcsin then To find the derivative of arccos d = Let g) = cos, 0 π. Its inverse is g ) = arccos, with graph shown below. arccos Graph of arccos

2 Once again, write g ) = arccos as y = arccos, to be read as "y is the angle whose cosine is "). Thus cos y = When we differentiate implicitly and re-arrange the resulting equation we get d = sin y = ± cos = ± Since the slope of the tangent at any point on the graph of arccos is negative, we must take the negative sign. Therefore if y = arccos then d = The derivative of arctan Let h) = tan, π π. Its inverse is h ) = arctan with graph shown below. arctany Graph of arctan Write h ) = arctan as y = arctan, to be read "y is the angle whose tangent is ". It follows that tan y =. Take derivatives implicitly to get sec y ) d = + tan y ) d = We have used the identity sec y = + tan y. Since + tan y = +, it follows that To find the derivative of ln d = + If we let w) = e, where is any real number, then its inverse w ), > 0, is called the natural logarithm function, and it is denoted by ln. More precisely, if w) = e then w ) = ln. Its graph is given below Graph of ln

3 As before write w ) = ln as y = ln. Then e y =. Differentiate implicitly and solve for the derivative. Since e y =, the result should be d = ) As pointed out above, ln is defined for positive values of. For < 0, we have to consider another function, namely g) = ln ). By the chain rule, its derivative is g ) = ) ) = Thus the derivative of ln ) is also. The two results may be stated as follows: We summarize all these results in a table: If h) = ln then h ) = Function arcsin arccos arctan ln Derivative + The chain rule etends to these functions as follows: Function arcsin ) Derivative ) Derivative of what is in ) arccos ) ) Derivative of what is in ) arctan ) ln Derivative of what is in ) + ) Derivative of what is in Eercise. Calculate the derivative of each function and simplify your answer as much as possible. Where they appear, a and b are constants. a) f) = ln b) g) = arcsin b c) u) = ln ) e d) h) = arccos e) w) = arcsin f) v) = arctan b g) f) = arcsin6) h) g) = ln9) 6 i) f) = 5 arctan j) v) = arctan ) k) f) = arcsin ) l) u) = arctan ) m) g) = ln n) h) = lnsin ) o) v) = ln + ) p) w) = ln + b) q) u) = 5 arcsin ) r) z) = lnb) s) h) = 5 arctan8) t) f) = arctan ) a b u) w) = arcsin ) v) f) = arcsin a + b) w) v) = ln +5) + ) z) = arcsin ) a b

4 . A lake polluted by coliform bacteria is treated with bacteria agents. Environmentalists estimate that t days after the treatment the number N of viable bacteria per milliliter will be given by Nt) = 0t ln t 0) 0. a) Find the number of viable bacteria per milliliter for t =, t =.5, and t = 7.5 days. b) Find and interpret the derivative of N for t = 5 days and t = 9 days.. Show that: a) If f) = arcsincos ) then f ) = b) If g) = cos arcsin ) then g ) = c) If w) = lnsec + tan ) then w ) = sec. d) If u) = ln sec ) then u ) = tan e) If h) = arccos ) then h ) = f) If f) = arctan b ) where b is a constant then f ) = b b +. The function f) = sec is one-to-one on 0, π) therefore we may define an inverse function g) = arcsec. a) Show that its derivative is g ) =, > b) You showed above that the derivative of arccos arccos ) or this is pure chance? Defend your answer. ) is also. Is arcsec related to 5. The function f) = csc is one-to-one on 0, π) therefore we may define an inverse function h) = arccsc. Determine its derivative. 6. Given h) = cot arcsin ), determine h ) and simplify your answer. 7. Given g) = ln csc ), determine g ) and simplify your answer. + ) ) 8. To find the derivative of a complicated function like f) = ln, first epand the + 5) logarithm epression: + ) ) ln = ln + ) + ln ) ln + 5 ) ) + 5) The derivatives of the epressions in the right hand side of ) are much easier to determine. We obtain f ) = Determine the derivative of each function below in a similar way: [ a) ln + sin ) ] + ) c) ln 5 b) ln + ) 5) + d) ln + ln 9. Let p be a positive number. Use L Hopital s rule to show that lim p = 0.

5 0. Assume that you are required to determine lim tan ) / π ) π. A direct substitution of π work because the eponent becomes infinite when = π. Let f) = tan )/ π ). Show that Now use L Hopital s rule to verify that lim π ln f)) = a) Determine lim csc π ) in a similar way. ln tan ) π. does not ln f)) =, then deduce the value of lim tan ) / π ) π.. Let n be a constant and f) = n ln,. By applying the Mean Value Theorem to f on the interval [, ], show that f) f) > 0 for all >. Use this to deduce that n > + ln for all >.. Suppose you are required to determine lim 0 + ln. In this form, it does not fall into one of the two ) ln forms to which we applied L Hopital s rule earlier on. However, if we write ln as, it falls ) ln into the second form. Use L Hopital s rule to determine. lim 0 + Now that we have the derivative of ln, we can determine derivatives of other functions involving eponents. Eample We know that the derivative of the eponential function u) = e, to base e is u ) = e. However, the derivatives of eponential functions to other bases are a little diff erent. For eample, consider f) =. Its derivative is not quite. To determine it, write y = then take logarithms of both sides to base e. The result is ln y = ln ) = ln Now that is no longer an eponent, we can easily take derivatives implicitly, with respect to, remember that ln is a constant), to get = ln. y d Since y =, solving for d gives f ) = d = ln ) y = ln ). You should be able to show, in a similar way, that if b is any positive number then the derivative of v) = b is v ) = ln b) b. This fact is worth recording: The derivative of b is ln b)b Eample To determine the derivative of g) = ), we write g) as y to get y = ) then take logarithms of both sides to base e. We end up epressing ln y as a product of two familiar functions: ln y = ln Taking derivatives of both sides, with respect to, gives y d = ln + = ln + = ln + ). It follows that g ) = = ln + ) y = ln + ) ). d 5

6 Eample Let b be a positive number that is not equal to. To determine the derivative of h) = log b, we use the change of base formula to write h in terms of the natural logarithm as It follows that h ) = ln b = ln b). h) = ln ln b Eercise 5. Here is another method of determining the derivative of f) = b : Since e is the inverse of ln, b = e ln b. Therefore f) = e ln b) = e ln b Now use the chain rule to determine f ).. Determine the derivative of each function: a) u) = 5 b) v) = log 0 + ) c) w) = + ) / d) f) = e) g) = ln +) +) +5) f) z) = ) + g) g) = ) h) h) = ) i) u) = ). Use the chain rule to find the derivative of a) f) = arctan b) g) = arcsin ) c) h) = arccos ). Let a be a constant. Show that the derivative of w) = ln + a + ) is w ) = a + ) a 5. Let a be a constant and v) = ln +. Show that v a ) = a The ph of a substance is defined by the formula ph = log [ H +] + a where [H + ] is the concentration of hydrogen ions in the substance, measured in moles per liter, which is abbreviated to moles/l). A certain chemical reaction causes the concentration of hydrogen ions in a substance to increase at the rate of 0.00 moles/l)/sec. Find and interpret the rate of change in ph when [H + ] = 0.5 moles/l. 7. When oyhemoglobin is reduced to hemoglobin, the hemoglobins are able to bind more H + ions in the blood thus regulating the acid level of blood. A formula to calculate a patient s blood ph level is ) ph = 6. + log, where is the base concentration and y is the acid concentration. During hyperventilation, oyhemoglobin is not reduced to hemoglobin quickly. The alveolar carbon dioide pressure decreases, blood base level decrease and blood acid level increases. This condition is known as respiratory alkalosis where the arterial blood ph level eceeds 7.5 a) Find the blood base to acid concentration ratio /y) for alkalosis. b) During alkalosis if d = 0.5 meq/l)/sec and = 0. meq/l)/sec when = 8 meq/l and dt dt y =.6 meq/l, what is the rate of change in ph? y 6

7 c) Interpret and eplain scenario b) above. d) How many moles per liter of hydrogen ions must be present for alkalosis to be diagnosed? 8. Acid rain is toic to vegetation and aquatic life. In northeastern U.S. forest soils are naturally acidic and their surface waters are mildly alkaline basic). Acid rain increases the amount of positive hydrogen ions in the soil. If rainfall over a lake causes the ph to decrease at the rate of dph = 0.8/hr, find dt the rate of change in hydrogen ions in the lake. 9. Give the most general antiderivative of the given function: a) f) = 5 d) h) = + ) g) w) = j) h) = + b) g) = 5 e) u) = + h) f) = k) u) = 5 + c) h) = ) f) v) = 5 + ) i) g) = + l) v) = 6 7