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Exponential and logarithmic functions are two important forms of a function. If the function is in the form of an expression that contains the variable in the exponent’s place, the function is called as an exponential function. Similarly, if the function is in the form of an expression in which the variable is a logarithm to any base, then it is called as a logarithmic function.

Exponential functions predict the growth or decay of the output with the change in variable. A logarithmic function depicts the change in function in terms of the power of a number.

For example, if y = log (x), then the function changes according to the power of 10 of the variable.The derivatives of exponential and logarithmic functions give an idea about the instantaneous rate of change of a function.

Exponential functions predict the growth or decay of the output with the change in variable. A logarithmic function depicts the change in function in terms of the power of a number.

For example, if y = log (x), then the function changes according to the power of 10 of the variable.The derivatives of exponential and logarithmic functions give an idea about the instantaneous rate of change of a function.

f(x) = ln (x). Let h be an infinitesimally small change in the variable, that is h is far less than x.

f(x + h) = ln (x + h)

Hence, f(x + h) – f(x) = ln (x + h) – ln (x) = ln [(x + h)/(x)] = ln [1 + ($\frac{h}{x}$)], applying logarithmic rules and simplifying.

As per the definition of a logarithmic series,

ln $[\left ( \frac{(x+h)}{(x)} \right )]$ = ln $[1+\left ( \frac{(h)}{(x)} \right )]$ = $\left ( \frac{h}{x} \right )$ - $\left ( \frac{h^2}{2x^2} \right )$ + $\left ( \frac{h^3}{2x^3} \right )$ - $\left ( \frac{h^4}{2x^4} \right )$ + ………

Therefore, [f(x + h) – f(x)]/[h] = $\left ( \frac{1}{x} \right )$ - $\left ( \frac{h}{2x^2} \right )$ + $\left ( \frac{h^2}{2x^3} \right )$ - $\left ( \frac{h^3}{2x^4} \right )$+ ………

Now, as per the definition of derivatives,

f ‘(x) = limit h -> 0 of [f(x + h) – f(x)]/[h] = limit h -> 0 of [$\left ( \frac{1}{x} \right )$ - $\left ( \frac{h}{2x^2} \right )$ + $\left ( \frac{h^2}{3x^3} \right )$ - $\left ( \frac{h^3}{2x^4} \right )$ + ………] = $\left ( \frac{1}{x} \right )$

Thus, the derivative of ln (x) = $\left ( \frac{1}{x} \right )$

We have found the above derivative. But what happens if the base of the logarithm is not the natural base but another constant say, b.

Thus, the derivative of ln (x) = $\left ( \frac{1}{x} \right )$

Let g(x) = log

But by change of the base rule of logarithms, g(x) = [ln(x)]/[ln (b)].

Therefore, g’(x) = 1/bx.

Now let us evaluate the derivatives of exponential functions.

Let, f(x) = a

Taking logarithm on both sides to the natural base,

ln [f(x)] = ln (ax)

or, ln [f(x)] = x*ln (a)

Now by implicit differentiation with respect to x,

$\left [ \left ( \frac{1}{f} \right )x \right ]*$ f’(x) = ln (a)

or, f’(x) = f(x) $\times$ ln (a)

or, f’(x) = a

Thus, the derivative of a

Even if the exponent is any other expression in x, the derivative can be found by using the chain rule.

As a special case, if a = e, the exponential constant, then, plugging in a = e, the derivative of ex = ex *ln (e) = ex * 1 = ex.

Thus, we get an important result that the derivative of ex = ex itself !

y = $e^{e^{x^{2}}}$ find the derivative of exponential function?

Given y= $ e^{e^{x^{2}}}$

Take log on both side

=> ln y = ln $e^{e^{x^{2}}}$

=> ln y = $e^{x^{2}}$ ln e

=> ln y = $e^{x^{2}}$

Differentiate with respect to x on both side

=> $\frac{y^{'}}{y}$ = $e^{x^{2}}$ $\times$ (2x)

=> y' = $e^{x^{2}}$ $\times$ (2x) $\times$ y

=> y' = (2x) $e^{x^{2}}$ $e^{e^{x^{2}}}$

y = $\frac{ln(2x)}{x^{5}}$ find the derivative of exponential function?

Given $\frac{ln(2x)}{x^{5}}$

=> y = x - 5 ln(2x)

use the product rule

$\frac{d(uv)}{dx}$ = (u $\times$ v' ) + (v $\times$ u')

=> y' = $x^{-5}$ $\times$ $\frac{1}{2x}$ $\times$ (2) + ln(2x) $\times$ $-5x^{-6}$

=> y' = $x^{-6}$ (1- 5 ln(2x))

=> y = $\frac{1-5 ln(2x)}{x^{6}}$

Consider an example of f(x) = log (x). The function is not defined at x = 0. Therefore, the y-axis is the vertical asymptote.

When x = 10, y = 1 and when x = 100, y = 2. So the compatible ordered pairs to plot on the grid are (10, 1)

and (100, 2). The graph is drawn as shown below:

Now, in case of exponential functions, a function changes far more rapidly as compared to the change in variable. Hence the graph of an exponential function will be horizontally shorter and more steep in vertical. Make a table of values that are compatible to plot the points on a grid. The y - intercept is a must. An exponential graph will always have a horizontal asymptote when the variable tends to $\pm\infty$ and the value of the function approaches a steady value. That will give an idea for the equation of the horizontal asymptote. Draw a smooth curve joining the points that are plotted taking the vertical asymptote as guidance.

Consider an example of f(x) = 2x. When the variable approaches $\infty$, the function tries to become 0. Therefore, the x-axis is the horizontal asymptote.

When x = 0, y = 1, when x = 1, y = 2, when x = 2, y = 4 and when x = 3, y = 8. The compatible ordered pairs to plot on the grid are (0, 1), (1, 2), (2, 4) and (3, 8). The graph is drawn as shown below: