Derivatives of Exponential functions of e

Derivative is the rate of change at a point which gives the slope of the curve at that point. When the given equation is y=f(x), the derivative is written as dy/dx or d[f(x)]/dx.  To find the derivatives of exponential functions, let us take a quick look at them. Exponential functions are the functions written in the form y = b^x, where b is a positive number that does not equal 1 and x is any real number.  They have a constant base and the exponent is a variable. The most important exponential function is e as the base, which is an irrational number. The function is written as, e(x) and is called the natural exponential function.  Now that we learnt about the natural exponential function e^x, let us learn more about the Derivatives of E.

The natural exponential function is remarkable and so are its derivatives. Let us first find the derivative of E, where E is f(x)=e^x :
As per the definition of derivatives, we get,
d[f(x)]/dx = lim(delta(x)?0) e^[x+delta(x)-e^x]/delta(x)
=lim(delta(x)?0) [e^xe^delta(x)- e^x]/delta(x)
=lim(delta(x)?0) e^x[e^delta(x)-1]/delta(x)
=lim(delta(x)?0) e^x[1+delta(x)-1]/delta(x)
= lim(delta(x)?0)e^xdelta(x)/delta(x)
=e^x
If f(x) = e^x then f’(x) = e^x. This means that slope is the same as the given function value or value of y for all the points on the graph. The other Derivatives of E or derivatives of e^x are as given below:
If u is a function of x, the derivative of an expression in the form e^u can be obtained and is given by d(e^u)/dx = e^u. du/dx

If an exponential function with base b is given, then the derivative of that expression is given by
d(b^u)/dx = b^u.ln b.du/dx

Let us take an example, derivative of E 2 which is derivative of e^x where x=2.  At this point x=2, the value of y=e^x  is approximately 7.39. We know that the derivative of e^x is e^x. So, the slope of the tangent, that is the derivative of e^2 at x=2 is also 7.39 approximately.

Derivative of E 2x will be the derivative of e^2x. To find the derivative of this exponential function, let us take y= E 2 or y= e^2x where u=2x. Using the chain rule, we get  dy/du = de^u/du , where du/dx equals 2.  So, d/dx of [e^2x] is (e^u). du/dx = 2. e^u , substituting u=2x, the derivative of e^2x is 2e^2x

Derivative of E 3x will be the derivative of e^3x. Using d(e^u)/dx = e^u. du/dx where u=3x, we get, e^3x. du/dx which will be 3e^3x as du/dx = 3

Law of Cosines Explained

Trigonometry Law of cosines:
Trigonometry is a field of study relating the angles and sides of a triangle. However, the fundamental ratios are derived easily from a right-angled triangle and are as identified Pythagorean ratios, yet the definite correspondence between the sides and angles can be established using the law of cosines. It is the relation between the sides and cosine of angle.


Convention:
All the angles are depicts upper case letters and sides are depicted by lower case letters. The side opposite to a vertex is represented by the corresponding lower case letter. In a triangle ABC, the side AB= c, BC = a, and AC = b.


Trigonometry law of cosines:
a^2= b^2+c^2 - 2bc cos (A)
b^2 = a^2 +c^2 – 2ac cos (B)
c^2 = a^2 + b^2 – 2ab cos (C)


Law of cosines Example Problems: 
For example, let us consider a triangle ABC, in which a= 3, b= 4 and c= 5.
To evaluate the angle C,
Substitute the values of a, b and c in
c^2 = a^2 + b^2 – 2ab cos (C)
5^2 = 3^2 + 4^2 – 2(3) (4) cos (C)
25= 9+16 – 24 cos(C)
25 = 25 – 24 cos(C)
Solving for cos (C), we get cos(C) =0. Hence, C = 90?.


Prove Law of Cosines:
We always rely on the principles of geometrical principles to prove the laws in trigonometry. To prove the law of cosines we use Pythagorean principle.

 In the above figure, CP is perpendicular to AC extended to P. Hence, BP = a sin(C) and CP = a cos(C).
Applying Pythagoras theorem in the right-angled triangle APB we get,
(AP)^2 + (BP)^2 = (AB)^2
(b – a sin(c))^2 + (a sin(C))^2 = c^2
Expanding        b^2 -2ab cos(C) + b^2 cos2(C) + b^2sin2(C) = c^2
b^2 +a^2 – 2ab cos(C) = c^2


Law of Cosines Problems
Law of cosines, in Physical sciences and technology, has very wide applications. The law of cosine gives us the magnitude of the difference of two vectors acting at an angle. For example, to evaluate the magnitude of the difference of two vectors of magnitude 100 units each acting at an angle of 120?, we get
c^2= 1002+1002 – 2(100) (100) cos (120)
c^2= 1002+1002 -1002
Solving for c, we get c= 100 units.

Derive Law of Cosines
Though there are so many methods by which one can prove the law of cosines, we stick to the use of Pythagorean principles.

In the above figure we have, when CP is perpendicular to AB,
c= a cos (B) + b cos (A)
c^2= ac cos (B) +bc cos (A)
Similarly
b^2=bc cos (A) + ba cos(C)
a^2= ab cos(C) + ac cos (B)
Adding the above two equations we have
b^2 + a^2 = ac cos (B) + bc cos (A) + 2 ab cos(C)
Comparing the above equation with (1), we get
b^2 + a^2 = c^2 + 2ab cos(C)
b^2 + a^2 -  2ab cos(C) = c^2

Derivatives and Graphs of Exponential and Logarithmic Functions

Exponential and Logarithmic Functions
The Logarithmic function with base b is a function, y = logb x. Here b is greater than zero and the function x is defined for all x greater than zero. An Exponential function with base b is a function, y=bx, defined for every real number.
Inverse Function: To find an inverse function (f-1), we need to interchange x and y and then solve for y. Example: f -1( x)  of 2x +1 will be, y=2x+1 (interchange x and y and solve for y)
x =2y+1
y = (x-1)/2 = f-1(x)
The Exponential functions and Logarithmic functions are inverse functions, that is, for any base b, the functions f(x) = logb x, g(x) = bx  are inverses.
Example: let f(x) =ln x and g(x) = ex then f and g satisfy the inverse functions. f(g(x) = ln ex=  x and g(f(x) = eln x= x, f(g(x) = g(f(x) and hence the functions f(x) and g(x) are inverses

Derivatives of Logarithmic and Exponential Functions
The most common exponential and logarithmic functions are natural exponent function ex, and the natural logarithm function, ln(x). The derivatives of Exponential and Logarithmic Functions are:
Exponential Functions derivative: d/dx (ex) = ex   d/dx(ax) =ax ln a
Logarithmic Functions derivative: d/dx(ln x) =1/x    d/dx(loga x)= 1/x ln a
Example: Derivative of f(x) = e3x+2  is given by d/dx (e3x+2) = e3x+2. 3 = 3e3x+2
   Derivative of f(x) = ln (3x+2) is given by d/dx[ln (3x+2)] = [1/(3x+2)]. 2 = 2/(3x+2)

Graphing Exponential and Logarithmic Functions
Exponential functions play a large role in real life. From science to money, graphing these exponential functions provide a visual representations to many applications in real life. Graphs of Exponential and Logarithmic functions using examples are as follows,
Let us now graph an exponential function, f(x) = 2x. First we evaluate f(x) using the integers -3, -2, -1, 0, 1,2,3 and tabulate the values.
x   -3          -2        -1        0 1 2        3
f(x)    1/8 1/4 1/2 1 2 4 8


(x,y)   (-3,0.125)      (-2, 0.25)   (-1,0.5)   (0,1)       (1,2)         (2,4)       (3,8)
Once we get the ordered pairs (x,y) plot the points which gives us the graph of the exponential function f(x)=2x.

Graphing Logarithmic Functions: There are several ways to graph logarithmic functions. The easiest way to graph them is to re-write them in exponential form.
Example:  Graph the logarithmic function, f(x) = log5 x. Re-writing f(x) = y =log5 x in exponential form we get x = 5y, choose values for y and then compute corresponding values for x.  Tabulating the values,
y   -1         0 1        2
x = 5^y 1/5         1             5       25
(x,y) (0.2,-1)     (1,0)           (5, 1)      (25, 2)
Plot the (x,y) values. The graph we get is the graph of the logarithmic function, f(x)= log5 x.