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It describes the rate of change in an exponential function with respect to the independent variable. The derivative of an exponential function like ax (a > 0) is the same function multiplied by a constant (the natural log of a). For example, ax becomes ax ln a when differentiated.
It can be obtained through the first principles of differentiation, utilizing limit formulas, and its graph increases when a > 1 and decreases when 0 < a < 1.
The formula for the differentiation or derivative of the exponential function is:
It can be proved by using the first principle, where we find the derivative of a function using the definition of limits. From the definition of limits, we know that
For the function f(x) = ax, using the First Principle,
=
=
= ax ln a {Since, }
Hence, the derivative of the Exponential Function ax is the product of ax and the natural logarithm of a.
Taking a = e in ax , our function will be f(x) = ex
So, f'(x) = ex ln e
ex ln(e) = ex logee (ln is a natural logarithm with base e)
ex logee = ex (from logarithm formulas, logaa = 1 )
Hence, f'(x) = ex
The nature of the graph of exponential function changes when the value of 'a' increases or decreases with respect to 1.
Since exponential function ax is only defined when a > 0 and derivative of a function is tangent to the graph or curve of that function. Hence, the graph of exponential function derivative is increasing for a > 0.
Exponential Function and Logarithmic Function are inverse of each other with change in base.
Exponential Form | Logarithmic Form |
|---|---|
x = ay | y = logax |
x = ey | y = ln x |
The derivative of exponential and logarithmic functions is mentioned below:
Example 1: Find the derivative of esin x.
Solution:
Given that y = esin x
using chain rule
dy/dx = esin x(cos x)
Example 2: Differentiate elog x.
Solution:
Let y = elog x
then dy/dx = elog x.1/x
Now we know that from the property of logarithm that elog x = x
Hence, dy/dx = elog x.1/x = x.1/x = 1
Example 3: Find the derivative of ex.log x.
Solution:
y = ex.log x
Here the function is in u.v form
Applying the product rule of differentiation we u.v' + v.u'
dy/dx = ex(log x)' + log x(ex)'
⇒ dy/dx = ex.1/x + log x.ex
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