f : R -> R : x -> a^{x}is called an exponential function with base a.
From the graphs we see that
y = 3^{x} ; y = 0.5^{x} ; y = 10^{0.2x-1}
f : R -> R : x -> a^{x}are either increasing or decreasing, the inverse functions are defined. The inverse function is called the logarithmic function with base a. We write
log_{a}(x) log_{10}(x) is written as log(x)So,
log_{a}(x) = y <=> a^{y} = x |
log_{2}(8) = 3 ; log_{3}(sqrt(3)) = 0.5 ; log(0.01) = -2From the definition it follows immediately that
for x > 0 we have a^{loga(x)} = x and for all x we have log_{a}(a^{x}) = x |
From the graphs we see that
log_{2}(x) ; log(2x+4) ; log_{0.5}(x)
Let log(x.y) = u then a^{u} = x.y (1) Let log(x) = v then a^{v} = x (2) Let log(y) = w then a^{w} = y (3) From (1) , (2) and (3) a^{u} = a^{v} . a^{w} => a^{u} = a^{v + w} => u = v + wSo,
log_{a}(x.y) = log_{a}(x) + log_{a}(y) |
log_{a}(x/y) = log_{a}(x) - log_{a}(y) |
log_{a}(x^{r} ) = r.log_{a}(x) |
log(x^{2} y^{3}) = 2 log(x) + 3 log(y) log(x^{2} / y^{3}) = 2 log(x) - 3 log(y) log( x^{y} )= y log(x) 1 log ------ = log(1) - log((xy)^{3}) = -3log(xy) = -3 (log(x)+log(y)) (xy)^{3} log(2x) + log(3x) = log(6x^{2}) 3log(x) + a log(x) = (3+a) log(x) = log(x^{3+a}) 0.5 log(x) = log(sqrt(x))Show that x^{log(y)} / y^{log(x)} = 1
The statement implies that x and y are positive, so we can write
x^{log(y)} / y^{log(x)} = 1 <=> x^{log(y)} = y^{log(x)} <=> log( x^{log(y)} ) = log( y^{log(x)} ) <=> log(y).log(x) = log(x).log(y)
1 log_{a}(x) =( -------) . log_{b}(x) log_{b}(a) |
log_{b}(a) . log_{a}(x) = log_{b}(x) Let log_{b}(a) = u then b^{u} = a (1) Let log_{a}(x) = v then a^{v} = x (2) Let log_{b}(x) = w then b^{w} = x (3) From (2) and (3) we have a^{v} = b^{w} Using (1) b^{u.v} = b^{w} So, u.v = w => log_{b}(a) . log_{a}(x) = log_{b}(x)
log(12.5) = 1.0969 log_{2}(12) = log(12)/log(2) = 3.58496 log(1/154)= -log(154) = -2.1875 log_{7}(0.5^{14}) = 14 log(0.5)/log(7) = -4.9869 log(-12.4) is not defined
f(x) = a^{x}Appealing on the definition of the derivative, we can write
(f(x+h)-f(x)) f'(x) = lim --------------- h->0 h a^{x+h} - a^{x} = lim ------------ h->0 h a^{x} (a^{h} - 1) = lim ----------- h->0 h (since a^{x} is constant relative to h ) (a^{h} - 1) = a^{x} . lim ----------- h->0 h Now, (a^{h} - 1) lim ----------- is a constant depending on the value of the base a. h->0 hIt can be proved that there is a unique value of a, such that this limit is 1. This very special value of a is called e.
(e^{h} - 1) lim ----------- = 1 h->0 h
(e^{h} - 1) lim ----------- = 1 0 hmeans that for very very small values of h
e^{h} - 1 is approximately h <=> e^{h} is approximately h +1 <=> e is approximately (1 + h)^{1/h}So,
e = lim (1 + h)^{1/h} = 2.718 28... 0 |
e = lim (1 + 1/t)^{t} = 2.718 28... infty |
log_{e}(x) = ln(x) |
The number a is a strictly positive number.
a^{k} = e^{r} <=> ln(a^{k}) = r <=> r = k.ln(a) Then (a^{k})^{x} = (e^{r})^{x} <=> r = k.ln(a) for all x <=> a^{kx} = e^{rx} <=> r = k.ln(a) for all xThe functions A.a^{kx} and A.e^{rx} are identical if and only if r = k.ln(a)
Example:
14 . 3^{2.7x} is identical with 14 . e^{2.97x} 6 . (0.25)^{-x} is identical with 6 . e^{1.39x}Working with the functions with base e, has many benefits for algebraic calculations.
The product of the functions with base e, from the previous example, is much easier to work with.
Compare the expressions: 14 e^{2.97 x} 6 e^{1.39 x} and 14 3^{2.7x} 6 (0.25)^{-x}.
We'll find the value of h, such that y_{2} = y_{1}/2.
y_{2} = (1/2) y_{1} <=> e^{-r(a+h)} = (1/2) e^{-ra} <=> e^{-ra} e^{-rh} = (1/2) e^{-ra} <=> e^{-rh} = (1/2) <=> - rh = ln(1/2) <=> - rh = ln(1) -ln(2) <=> - rh = -ln(2) <=> h = ln(2)/rIf we increase x with ln(2)/r, then the image of e^{-rx} reduces to half the initial value. Remarkable is that ln(2)/r does not depend on our initial value a.
This means that the image is halved, when a random initial value of x increases with ln(2)/r.
That's why ln(2)/r is called the 'half value' of the function e^{-rx}.
Example : Take the function y = e^{-0.5x}.
The half value is ln(2)/0.5 = 1.386
Exercise: plot the graph and check all previous properties.
Let f(x) = log(x) , then (f(x+h)-f(x)) f'(x) = lim --------------- h->0 h (log(x+h)-log(x)) <=> f'(x) = lim ------------------- h->0 h log( (x+h)/x ) <=> f'(x) = lim ------------------- h->0 h 1 <=> f'(x) = lim --- . log( (x+h)/x ) h->0 h <=> f'(x) = lim log( (x+h)/x )^{1/h} h->0 <=> f'(x) = lim log( (x+h)/x )^{1/h} h->0 <=> f'(x) = lim log(1 + h/x)^{1/h} h->0 <=> f'(x) = lim log((1 + h/x)^{x/h} )^{1/x} h->0 <=> f'(x) = lim (1/x).log(1 + h/x)^{x/h} h->0 <=> f'(x) =(1/x). lim log(1 + h/x)^{x/h} h->0 <=> f'(x) =(1/x). lim log(1 + h/x)^{x/h} h/x->0 <=> f'(x) =(1/x).log lim (1 + h/x)^{x/h} h/x->0 <=> f'(x) =(1/x).log(e) <=> f'(x) =(1/x).ln(e)/ln(a) <=> f'(x) =(1/x)/ln(a) 1 <=> f'(x) = ---------- x. ln(a)
d 1 -- log_{a}(x) = ---------- dx x. ln(a) d 1 -- log_{a}(u) = ---------- . u' dx u. ln(a) d 1 -- ln(x) = --- dx x d 1 -- ln(u) = ---.u' dx u |
y = ln(2x^{2}+6) y' = (1/(2x^{2}+6)). 4x --------------------------- y = ln^{2}(x) y' = 2 ln(x) . (1/x) --------------------------- y = ln(1/x) y'= x.(-1/x^{2}) = -1/x --------------------------- y = ln(ln(x)) y'= (1/ln(x)) . (1/x) --------------------------- y = x^{3} ln(x) y'= (3x^{2}).ln(x) + (x^{3}).(1/x) = (3x^{2}).ln(x) + x^{2} --------------------------- y = log_{3}(x^{2}) y'= (1/(x^{2}.ln(3)).(2x) = 2/(x ln(3)) --------------------------- y = ln(x)/x x.(1/x) - ln(x) 1 - ln(x) y'= ------------------ = ------------ x^{2} x^{2} --------------------------- y = ln(sqrt(2x^{2}+x)) first we rewrite y y = 0.5 ln(2x^{2}+x) 4x+1 y'= 0.5 ---------- 2x^{2}+x --------------------------- y = sqrt(ln(x^{2})) 1 y'= --------------- .(2/x) 2 sqrt(ln(x^{2}))
Let f(x) = a^{x}, then log_{a}(a^{x} ) and x are identical functions. Hence, the derivative of both functions is the same. So, 1 ---------- .(a^{x} )' = 1 a^{x} .ln(a) d <=> ---(a^{x} ) = a^{x} .ln(a) dx
d ---(a^{x} ) = a^{x} .ln(a) dx d --(e^{x} ) = e^{x} dx d --(a^{u} ) = a^{u} .ln(a).u' dx d --(e^{u} ) = e^{u} .u' dx |
y = x.e^{-5x} y'= e^{-5x} + x.(-5)e^{-5x} --------------------------- y = 3^{2x} y'= 3^{2x} ln(3) 2 = 3^{2x} ln(9) --------------------------- y = e^{x}/x^{2} x^{2} e^{x} - e^{x} 2x y'= ------------------ x^{4} e^{x} (x-2) = ------------ x^{3} --------------------------- y = (3e)^{x} y'= (3e)^{x} ln(3e) = (3e)^{x} (ln3 +lne) = (3e)^{x} (ln3 + 1) --------------------------- y = arcsin(2^{x}) 2^{x} ln(2) y'= ----------------- sqrt(1-2^{2x})
Let f(x) = x^{r} with r any real number. x^{r} = e^{r.ln(x)} => d --(x^{r}) = e^{r.ln(x)}.(r.ln(x))' dx = x^{r}.r.(1/x) = r.x^{r-1}Thus,
For any real number r, we have
d --(u^{r}) = r.u^{r-1}.u' dx |
u^{v} = e^{v.ln(u)} d --(u^{v}) = e^{v.ln(u)}.(v.ln(u))' dx = u^{v} . (v' ln(u) + v.(1/u).u' = v u^{v-1} u' + u^{v}.ln(u).v'
d --(u^{v}) = v u^{v-1} u' + u^{v}.ln(u).v' dx |
y = x^{x} y'= x^{x} ln(x) + x x^{x-1} = x^{x} (1 + ln(x)) y = (ln(x))^{2x} y'= (ln(x))^{2x} ln(ln(x)).2 + 2x.(ln(x))^{2x-1} (1/x) y = (e^{x})^{x} we rewrite y: y = e^{x2} y'= e^{x2} 2x
ln(x) 1/x lim ------- = lim ------- = 0 infty x infty 1
x^{n} n.x^{n-1} lim ------ = lim --------- infty e^{x} e^{x} n.(n-1)x^{n-2} = lim -------------- e^{x} = ... n! = lim ------ = 0 e^{x}
ln(x) 1/x lim xln(x) = lim ------ = lim ------ = - lim x = 0 0 1/x -1/x^{2}
lim x^{x} = lim e^{x.ln(x)} = e^{lim x.ln(x)} 0 0 and from previous example = e^{0} = 1
lim (cos(x))^{1/x} 0 = lim e^{(1/x).ln(cos(x))} 0 = e^{lim (1/x).ln(cos(x))} but lim (1/x).ln(cos(x)) 0 ln(cos(x)) = lim --------------- 0 x -sin(x)/cos(x) = lim ---------------- = 0 0 1 So, lim (cos(x))^{1/x} = e^{0} = 1 0
lim ( x ln(e + 1/x) -x ) infty = lim x ( ln(e + 1/x) - 1) ln(e + 1/x) - 1 = lim ----------------- 1/x -1/x^{2} = lim ------------------- (e + 1/x) . (-1/x^{2}) = 1/e
We calculate the following limit for x --> + infinity
lim x^{1/x}First we calculate
lim ln( x^{1/x} ) = lim (1/x) ln(x) = lim ln(x)/x ( case infinity / infinity) de l'Hospital = lim 1/x = 0The initial limit is e^{0} = 1
We calculate the following right limit for x --> 0
lim ( tan(x) )^{x}First we calculate
lim ln ( tan(x) )^{x} = lim x.ln(tan(x)) ( case 0. infinity ) ln (tan(x)) = lim -------------- ( case infinity / infinity) 1/x de l'Hospital - x^{2} = lim -------------------- tan(x) cos^{2}(x) x^{2} = - lim ---------- tan(x) x = - lim --------- . x = 1.0 = 0 tan(x)The initial limit is e^{0} = 1
We calculate the following right limit for x --> 0
lim sin(x)^{sin(x)}First we calculate
lim ln sin(x)^{sin(x)} = lim sin(x) ln(sin(x)) ( case 0.infinity ) ln(sin(x) = lim ---------------- ( case infinity / infinity) ( 1/ sin(x)) de l'Hospital ( 1/ sin(x)).cos(x) = lim ------------------------- (-1/ sin^{2}(x)) . cos(x) ( 1/ sin(x)) = lim ----------------- (-1/ sin^{2}(x)) = - lim sin(x) = 0The initial limit is e^{0} = 1