Question

Find the derivative of $ \log ({e^x}) $ using first principle.

Answer 1

Hint: In the question they have clearly mentioned to use the first principle of differentiation which is stated as follows,
Given a function $ y = f(x) $ , its derivative or the rate of change of $ f(x) $ with respect to $ x $ is defined as
 $ \dfrac{d}{{dx}}f(x) = f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h} $
Where $ h $ is an infinitesimally small positive number.

Complete step by step answer:
 Let $ f(x) = \log ({e^x}) $
Let us differentiate $ f(x) $ with respect to $ x $ by using the formula $ \dfrac{d}{{dx}}f(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h} $
For finding $ f(x + h) $ we replace $ x $ by $ x + h $ in the given function.
 $ \Rightarrow \dfrac{d}{{dx}}\log ({e^x}) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\log ({e^{x + h}}) - \log ({e^x})}}{h} $
From the properties of logarithmic function we know that logarithm of a number containing power can be written as $ \log ({x^m}) = m\log (x) $ .
Applying this property to our function we get, $ \dfrac{d}{{dx}}\log ({e^x}) = \mathop {\lim }\limits_{h \to 0} \dfrac{{(x + h)\log e - x\log e}}{h} $
If the base of logarithm is not mentioned explicitly it is understood that the base is $ e $ and logarithm of a number which is same as the base is always $ 1 $ , that is $ {\log _b}b = 1 $ which implies that $ {\log _e}e = 1 $ .
Substituting this in the above equation we get,
 $ \dfrac{d}{{dx}}\log ({e^x}) = \mathop {\lim }\limits_{h \to 0} \dfrac{{x + h - x}}{h} $
On cancelling $ x $ with opposite signs we have,
 $ \dfrac{d}{{dx}}\log ({e^x}) = \mathop {\lim }\limits_{h \to 0} \dfrac{h}{h} = \mathop {\lim }\limits_{h \to 0} 1 = 1 $ .
Therefore the derivative of $ \log ({e^x}) $ is $ 1 $ .

Note: In the question if they do not mention using first principle, we can use direct method to differentiate the function which is easier than first principle. Also the function here contains logarithmic function and when dealing with logarithmic functions all the logarithmic properties must be clearly known, then solving the problem becomes simple.