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.
Derivatives of Logarithmic and Exponential Functions
Let us start with finding the derivative of logarithmic functions. Assume the simplest form of an logarithmic function f(x) = ln (x). Even if the function is more descriptive, the derivative can still be figured out by using differentiation rules.
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) = logb (x)
But by change of the base rule of logarithms, g(x) = [ln(x)]/[ln (b)].
Therefore, g’(x) =$ \left [ \left ( \frac{1}{x*inb} \right ) \right ]$, since ln (b) is a constant.
Thus, the derivative of logb (x) = $ \left [ \left ( \frac{1}{x*inb} \right ) \right ]$.
Now let us evaluate the derivatives of exponential functions.
Let, f(x) = ax. (Again we have considered the basic form of an exponential function).
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)*ln (a)
or, f’(x) = ax *ln (a)
Thus, the derivative of ax = ax *ln (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 !
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) = logb (x)
But by change of the base rule of logarithms, g(x) = [ln(x)]/[ln (b)].
Therefore, g’(x) =$ \left [ \left ( \frac{1}{x*inb} \right ) \right ]$, since ln (b) is a constant.
Thus, the derivative of logb (x) = $ \left [ \left ( \frac{1}{x*inb} \right ) \right ]$.
Now let us evaluate the derivatives of exponential functions.
Let, f(x) = ax. (Again we have considered the basic form of an exponential function).
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)*ln (a)
or, f’(x) = ax *ln (a)
Thus, the derivative of ax = ax *ln (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 !
Derivatives of Logarithmic and Exponential Functions Examples
Here are some examples for derivatives of logarithmic and exponential functions
Solved Examples
Question 1: Find the derivative of the logarithmic function f(x) = log2 (5x2 + 3)
Solution:
Let f(x) = g(u), where g(u) = log2 (u) and u(x) = (5x2 + 3)
So, as per chain rule, f’(x) = g’(u)*u’(x)
g’(u) = $\frac{1}{u}$ = 1/[(5x2 + 3)*ln (2)] and u’(x) = 10x
Therefore, f'(x) = ${\frac{1}{[5x^2 + 3)*ln(2)]}}$ * (10x) = ${\frac{10x}{[5x^2 + 3)*ln(2)]}}$
Solution:
Let f(x) = g(u), where g(u) = log2 (u) and u(x) = (5x2 + 3)
So, as per chain rule, f’(x) = g’(u)*u’(x)
g’(u) = $\frac{1}{u}$ = 1/[(5x2 + 3)*ln (2)] and u’(x) = 10x
Therefore, f'(x) = ${\frac{1}{[5x^2 + 3)*ln(2)]}}$ * (10x) = ${\frac{10x}{[5x^2 + 3)*ln(2)]}}$
Question 2: Find the derivative of the exponential function f(x) = 3(2x – 5)
Solution:
Let f(x) = g(u), where g(u) = 3(u) u(x) = (2x – 5)
So, as per chain rule, f’(x) = g’(u)*u’(x)
g’(u) = 3(u) )*ln (3)= 3(2x - 5) )*ln (3) and u’(x) = 2
Therefore, f’(x) = 3(2x - 5) )*ln (3)*(2) = (2)* 3(2x - 5) )*ln (3)
Solution:
Let f(x) = g(u), where g(u) = 3(u) u(x) = (2x – 5)
So, as per chain rule, f’(x) = g’(u)*u’(x)
g’(u) = 3(u) )*ln (3)= 3(2x - 5) )*ln (3) and u’(x) = 2
Therefore, f’(x) = 3(2x - 5) )*ln (3)*(2) = (2)* 3(2x - 5) )*ln (3)
Graphing Exponential and Logarithmic Functions
As explained earlier, a logarithmic function changes far less compared to the change in a variable. Hence the graph of a logarithmic function will be horizontally broader and less steep vertically. Make a table of values that are compatible to plot the points on a grid. A logarithmic graph will always have a vertical asymptote for the value of the variable which makes the function undefined. Draw a smooth curve joining the points that are plotted taking the vertical asymptote as guidance.
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:
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: