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Exponential Functions solved questions
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CHAPTER 24Exponential Functions
24.1 Evaluate e '"'.
«-'»' = *'" <"-> = !/*.
24.2 Evaluate In e *.
In e ": = — x by virtue of the identity In e" = a.
24.3 Find(e 2 )' n *.
)'"- = ( ")2 = jc2. Here, we have used the laws 0")" = *"" and eln " = u.
24.4 Evaluate (3e)ln'.
S* \ln A- _ ^ ln3e\tn^M/ln3+l\lnAr_ /-'" Jr\ln3 + l __ In 3+1
24.5 Evaluate e1'"".
e1-tn' = e1e-la* = e/el'l* = e/x.
24.6 Findln(e*/:t).
In (eA/x) = In e* — In x = x - In x. We have used the identities In (u/u) = In u - In f and \ne" = u.
In Problems 24.7-24.16, find the derivative of the given function.
24.7 e '.
By the chain rule, Df(e *) - e'* • Dx(-x) = e *-(-\)=-e *. Here, we have used the fact that
Du(e") = e".
24.8 e1".
By the chain rule, D,(«l") = «"'' °,d^) = «"' ' (~1^ 2 ) = ~e"Vx2.
24.9 ecos *
By the chain rule, Dx(emx) = ecos" • Dx(cosx) = e0** • (-sin x) = -e cos * sin*.
24.10 tan e".
By the chain rule, D,(tan e') = sec2 e" • Dx(e") = sec2 e' • e' = e* sec2 e".
24.11 e'/x.
By the quotient rule,
24.12 e" In x.
By the product rule, Dx(e' In x) = e" • D,(ln x) + In jc • Dx(e") = e* • + In x • e' = e*1
24.13 xw.
Dx(x") = D,(e" '"') = e"'"* • Dt(ir In x) = e"ln* [In like manner, D,(xr) =
{
rx" for any real number r.]
195
, 196 CHAPTER 24
24.14 •n*.
D,(ir') = Ds(e''"*) = e'lnir• D,(x In IT) = e"ln" • In TT = In ir • TT*.
24.15 In e2'.
D,(lne 2 *) = D,(2*) = 2.
24.16 eA - e".
Df(e* - e~') = Dx(e") - Dx(e'") = e" - (-e'x) = e" + e'\ Here, Dx(e~*) = -e~" is taken from Prob-
lem 24.7.
In Problems 24.17-24.29, evaluate the given antiderivative.
24.17 J e3' dx.
Let u = 3x, du = 3dx. Then j e3" dx = !> $ e" du = %e" + C = ^e** + C.
24.18 J e-' <fc.
Let « = -*, du = -dx. Then / e"* rfx= -J e"rfw= -e" + C= -e "' + C.
24.19 $e*Ve^2dx
Let u = e'-2, du = e'dx. Then J e'^e* -2dx = f «"2 du = §w 3 ' 2 + C= i(\V -2) 3 + C.
24.20 fe c o s i sinA:djc.
J ecos' sin x dx = -ecos * + C, by Problem 24.9.
24.21 Ja'dx, for a ^ l .
a* = <•* ln °. So, let M = (In a)^:, dw = (In a) dr. Then
24.22 S32*dx,
Choose a = 32 = 9 in Problem 24.21:
24.23 J <T dx.
Let M = ax, du = a dx. Then
24.24 /V?djc.
by Problem 24.23.
24.25 /Ar"djc.
This is a special case of the general law for any
constant
24.26 J e'e2" dx.
J eV <b = J e'+2x dx = Se3'dx= ^e3' + C.
24.27
Let a = e* + l, du = e'dx. Then, noting that u > 0,
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