Download The Exponential Function - Practice Problems with Solution | M 408C and more Study notes Mathematics in PDF only on Docsity!
M408C: The Exponential Function
November 13, 2008
We define the natural exponential function as the inverse of the natural logarithm function:
We can do this since ln x is an increasing function, so it is one-to-one, and thus has an inverse. The two functions undo each other, i.e.
The domain of the exp is the range of ln, i.e. (−∞, ∞). The range of exp is the domain of ln, i.e. (0, ∞). Since ln(ex) = x ln(e) = x and exp is the inverse of ln, we have that ex^ = exp(x). This is the notation we will typically use for exp(x). The cancellation laws then become eln^ x^ = x for x > 0 and ln(ex) = x for all x. The limiting behavior of ex^ is well understood:
We also have the laws of exponents which follow from the laws of logarithms: ex+y^ = exey, ex−y^ = e
x ey^ , and (e
x)r (^) = exr. Finally, differentiating and integrating the exponential is straightfor-
ward:
For the general exponential function, we build on top of our previous work. (We will discuss the general logarithm function on Tuesday.) So if a > 0 and r is any number, then ar^ = (eln^ a)r^ = er^ ln^ a. Thus we can define ax^ = ex^ ln^ a. This is the exponential function with base a. The laws of exponents hold for ax^ as well: ax+y^ = axay, ax−y^ = a
x ay^ , (a
x)y (^) = axy, and (ab)x (^) = axbx. We also
have the derivative and integral formulas:
- (7.3*.48) Differentiate the function f (t) = sin^2 (esin
(^2) t ). Solution: We apply the chain rule along multiple times to get
f ′(t) = 2 sin(esin
(^2) t ) ·
[
sin(esin
(^2) t )
]′
= 2 sin(esin
(^2) t ) cos(esin
(^2) t ) ·
[
esin
(^2) t]′
= 2 sin(esin
(^2) t ) cos(esin
(^2) t )esin
(^2) t ·
[
sin^2 t
]′
= 2 sin(esin
(^2) t ) cos(esin
(^2) t )esin
(^2) t · 2 sin t cos t = 4 sin(esin
(^2) t ) cos(esin
(^2) t )esin
(^2) t sin t cos t
- (7.3*.82) Evaluate the integral
∫ (^) e 1 /x x^2 dx. Solution: We do a u-substitution by setting u = 1/x = x−^1 , so du = −x−^2 dx = − dxx 2 : ∫ e^1 /x x^2
dx = −
eudu = −eu^ + C = −e^1 /x^ + C.
- (7.4*.42) Differentiate the function y = (ln x)cos^ x.
Solution: This is a tricky problem. Although we know how to differentiate a number raised to a function – i.e. ax^ – we don’t know how to differentiate a function raised to a function. Instead, let’s try logarithmic differentiation – it will take the cos x and turn it from an exponent to a factor.
ln y = ln ((ln x)cos^ x) = cos x · ln (ln x).
Implicitly differentiating both sides, we get
y′ y
= − sin x · ln (ln x) + cos x ·
[ln x]′ ln x
= − sin x · ln (ln x) + cos x ·
1 /x ln x
= − sin x · ln (ln x) +
cos x x ln x
Solving for y′^ and substituting in for y, we get
y′^ = (ln x)cos^ x
cos x x ln x
− sin x · ln (ln x)
- (7.4*.50) Evaluate the integral
∫ (^2) x 2 x+1 dx. Solution: We use u-substitution with u = 2x^ + 1, so du = 2x^ ln 2 dx: ∫ 2 x 2 x^ + 1
dx =
u
du ln 2
ln 2
du u
ln 2
(ln |u| + C) =
ln 2
(ln | 2 x^ + 1| + C).
Since 2x^ + 1 > 0 for all x, | 2 x^ + 1| = 2x^ + 1. We can also multiply through by (^) ln 2^1 and rename (^) ln 2C as just C (in the end, it’s just a constant):
∫ 2 x 2 x^ + 1
dx =
ln(2x^ + 1) ln 2
+ C.
