The method that is used to differentiate functions by employing the logarithmic derivative of a function it is called logarithmic differentiation or differentiation by taking logarithms. The technique is commonly used in cases where it is easier to differentiate the logarithm of a function rather than the function itself. This generally occurs in cases where the function of interest is composed of a product of a number of parts; so that a logarithmic transformation will turn, it into a sum of separate parts (which much simpler to differentiate). It can also be useful when it is applied to functions raised to the power of variables or functions. Logarithmic differentiation depends on the chain rule as well as properties of logarithms ( in particular , the natural logarithm or the logarithm  to the base e) to transform products into sums and divisions into subtractions. The principle can be implemented, at least in part, in the differentiation of almost all differentiable functions, providing that these functions are non-zero.

The derivative of a function is the rate at which the function value is changing, with respect to x, at a given value of x. Therefore, if we can find the derivative of the distance function of the winner of any race, then it will be possible to find out how fast the winner is going at any given time in the race.

The difference quotient of a function f(x) is a formula that gives the slope of the line through nay points with x-coordinates and x+h on the function.

(f(x+h)-f(x)) I h

This is the key formula to calculate derivative, derivatives are computed by finding the limit of the difference quotient of a function as h approaches.

Basically, we can compute the derivative of  f(x) using the limit definition of derivatives with the following steps

1. Find f(x+h)
2. Plug f(x+h), f(x), and h into the limit definition of a derivative.
3. Simplify the difference quotient
4. Take the limit, as h approaches 0, of the simplified difference quotient.

Lets take a example to understand the racing functions f(x)=-7*2 + 280x. First it is required to find f(x+h)

f(x+h)=-7(x+h)² +280(x+h)= -7(x² +2xh+h²)+280x+280h =-7x²-14xh-7h²+280x+280h

From this it will be possible to calculate the winners speed at any time during the race . for instance , conceder his speed after 10 seconds . it is required to plug x=10 into the derivative formula

f(x)=-14(10)+280=140

we get that the derivative of fat x =10 is 140 , so at 10 seconds into the race he was driving at a speed of 140 mile per hour.

Another example of using the limit definition to compute a derivative . it is required to consider the function g(x) =1/x where x is not equal to 0 . to find the derivative using the limit definition of derivative , we first find the g(x+h).

g(x+h)= 1/(x+h)

now by plugging g(x+h) , g(x) and h into limit definition and find the limit.

It is monotonous to calculate a limit every time it is require to know the derivative of a function. Fortunately, it will be possible to develop a small collection of examples and rules that allow computing the derivative of almost any function. Many functions involve quantities raised to a constant power, such as polynomials and more complicated combinations like y= (sin x)⁴. Therefore it can be started by examining powers of a single variable, this will give a building block for more complicated examples.

If we start with the derivative of a power function, f(x) =xⁿ. Here n is the number of any kind of integer, rational, positive, negative, even irrationals. The simple examples are already been done so this formula is a complicated one

d/dx*xⁿ=nx^n-1

It is very complicated show this is true for any n. the easiest and most simple case is that n is a positive integer. To compute the derivative it is required to compute the following limit

d /dx*x n = lim ∆x→0 (x + ∆x) n − x n/ ∆x .

For a specific fair value of n , it can be done by using straight forward algebra .

Find the derivative of f(x) = x 3.

d /dx*x 3 = lim ∆x→0 (x + ∆x) 3 − x 3 /∆x .

 = lim ∆x→0 x 3 + 3x 2∆x + 3x∆x 2 + ∆x 3 − x 3/ ∆x .

= lim ∆x→0 3x 2∆x + 3x∆x 2 + ∆x 3 /∆x .

= lim ∆x→0 3x 2 + 3x∆x + ∆x 2 = 3x 2 .

The general case is not much complicated as long as we try to do too much . the main factor is understanding what happens when (x+∆x)ⁿ is multiplied out

(x + ∆x) n = x n + nxn−1∆x + a2x n−2∆x 2 + · · · + +an−1x∆x n−1 + ∆x n

We know that multiplying out will give large number of terms all of the form xⁱ∆x^j and in fact that i+j =n  in every term . one way to see this is to understand that one method for multiplying out (x+ ∆x)ⁿ is the following : in every (x+∆x) factor , take either the x or the ∆x, then multiply the n choices together , do this in all possible ways , for instance (x+ ∆x)³ there are eight possible methods  to do this

(x + ∆x)(x + ∆x)(x + ∆x) = xxx + xx∆x + x∆xx + x∆x∆x

 + ∆xxx + ∆xx∆x + ∆x∆xx + ∆x∆x∆x

= x 3 + x 2∆x + x 2∆x + x∆x 2

+ x 2∆x + x∆x 2 + x∆x 2 + ∆x 3

 = x 3 + 3x 2∆x + 3x∆x 2 + ∆x 3

It is not a factor what n denotes, there are n ways to pick ∆x in one factor and x in the remaining n-1 factors , this means one term is nx^n-1  . the other co efficient are somewhat difficult to understand , but these are not required therefore in this formula above they have simply been called a_2,a₃ and so on . it is known that every one of these terms contains ∆x to at least  the power 2. So the

d /dxx n = lim ∆x→0 (x + ∆x) n − x n/ ∆x

= lim ∆x→0 x n + nxn−1∆x + a2x n−2∆x 2 + · · · + an−1x∆x n−1 + ∆x n − x n /∆x

= lim ∆x→0 nxn−1∆x + a2x n−2∆x 2 + · · · + an−1x∆x n−1 + ∆x n/ ∆x

= lim ∆x→0 nxn−1 + a2x n−2∆x + · · · + an−1x∆x n−2 + ∆x n−1 = nxn−1.

Now without any problem it will be possible to verify the formula for negative integers this can be checked with the given example

Find the derivative of y = x −3. Using the formula, y ′ = −3x −3−1 = −3x −4.

 Here is the general computation. Suppose n is a negative integer; the algebra is easier to follow if we use n = −m in the computation, where m is a positive integer.

  d /dxx n = d dxx −m = lim ∆x→0 (x + ∆x) −m − x −m /∆x

= lim ∆x→0 1 (x+∆x)m − 1 xm /∆x

= lim ∆x→0 x m − (x + ∆x) m/ (x + ∆x)mxm∆x

= lim ∆x→0 x m − (x m + mxm−1∆x + a2x m−2∆x 2 + · · · + am−1x∆x m−1 + ∆x m) (x + ∆x)mxm∆x

= lim ∆x→0 −mxm−1 − a2x m−2∆x − · · · − am−1x∆x m−2 − ∆x m−1 ) (x + ∆x)mxm

 = −mxm−1 xmxm = −mxm−1 x 2m = −mxm−1−2m = nx−m−1 = nxn−1 .

Later we will check how the other cases of the power rule work. However, from now on we will use the power rule whenever n is any real number. Let’s note here a simple case in which the power rule applies, or almost applies, but is not really needed. Let assume that f(x)=1 has a graph that  this 1 is a function and not a simple number , and that f(x)=1 has graph that is a horizontal line , with zero slope. So from that it can be said that f(x)=0. This can also be written as f(x)=x⁰, though there is some question about just what this means at x=0 . If the power rule is applied we get f(x)=0x^-1=0/x=0,again noting that there is a problem at x=0.so the power rule works in this case, but its really best to just remember that the derivative of any constant function is zero.

Linearity of the derivative

An operation is linear if it behaves nicely with respect to multiplication by a constant addition. The name comes from the equation of a line through the origin, f(x)=mx , and the following two properties of this equation. First f(cx)=m(cx)=c(mx)=cf(x), so the constant c can be moved outside or moved through the function f. second f(x+y)=m(x+y)=mx+my =f(x)+f(y) so the addition symbol likewise can be moved through the function. The corresponding properties for this derivative are

(cf(x))′ = d dxcf(x) = c d dxf(x) = cf′ (x),

And

(f(x) + g(x))′ = d dx(f(x) + g(x)) = d dxf(x) + d dxg(x) = f ′ (x) + g ′ (x).