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Derivative of Logarithmic Function?


Differentiation is a process of finding the derivative is known as differentiation. Generally the differential coefficient of the function f(x) with respect to x is denoted by df(x)/dx or f’(x).

The process of finding differential coefficient by the definition of differentiation is called differentiation by first principle. It can be easily understood in the following steps.

1. At first write the function f(x) equal to x.

2. Then convert x to x+δx and y to y+δy.

3. Means y+δy =f((x+δx.)

4. Now evaluate the value of δy , say δy=f(x+δx)-f(x) .

5. Now in next step divide both the sides by δx, or δy/δx = f(x+δx)-f(x)/δx .

6. Find the limit of δy/δx when δx →0 .

7. So we can say dy/dx = lim δx →0 f(x+δx) -f(x)/δx.

8. Thus we can calculate derivative of any function.

So on the basis of above steps Derivative Of Logarithmic Function can be evaluated.

Let y = logex (taking the function equal to y)

Y+δy =log((x+δx)) (increment of δx in x and δy in y)

δy = log((x+δx)/x) (evaluating the value of δy by subtraction)

δy=log(1+ δx/x) (on simplifying )

δy= δx/x – ½( δx/x) 2 + 1/3( δx/x)3-……… (Expanding with the formula of log x)

δy/δx= δx/ δx [1/x – ½ δx/x2 +1/3 (δx)2/ x3-……] (dividing by δx both the sides)

lim δx →0 δy/δx = lim δx →0 [1/x – ½ δx/x2 +1/3 (δx)2/ x3-……] (on applying the limits)

dy/dx = 1/x – 0 +0 ….. (by the reference of Point no 7)

dy/dx = 1/x (on simplifying)

d(logex)/dx= 1/x (here the answer)

So by above description we can say that derivative of any function can be calculated with the help of differentiation by first principle. Thus the derivative of log x can be evaluated very easily.