20. Derivative of Logarithmic and Exponential Functions

Dated: 30-10-2024

Derivative of Logarithmic Functions

There is a special irrational constant which is defined as:

\[e = \sum_{n = 0}^\infty \frac{1}{n!}\]

We can alternatively define it in terms of limits.¹

\[e = \lim_{x \to 0} (1 + x)^{\frac{1}{x}} = \lim_{x \to 0}\left(1 + \frac{1}{x}\right)^x = 2.71\ldots\]

Let us define a function² such as

\[y = \log_b(x)\]

Then it can be differentiated as:

\[\frac{dy}{dx}=\frac{d}{dx}[log_{b}(x)]\]

\[= \lim_{h\to0}\frac{log_{b}(x+h)-log_{b}(x)}{h}\]

Using the theorems of logarithms³

\[=\lim_{h\to0}\frac{1}{h}log_{b}\left(\frac{x+h}{x}\right)\]

\[=\lim_{h\to0}\frac{1}{h}log_{b}\left(1+\frac{h}{x}\right)\]

Then we can define \(v = \frac{h}{x}\). Substituting it, we get:

\[=\lim_{v\to0}\frac{1}{vx}log_{b}(1+v)\]

Since the limit¹ is in terms of \(v\) now, therefore, the \(x\) is now treated as a constant.

\[=\frac{1}{x}\lim_{v\to0}\frac{1}{v}log_{b}(1+v)\]

Using the theorems of logarithms³ again, we get:

\[=\frac{1}{x}\lim_{v\to0}log_{b}(1+v)^{\frac{1}{v}}\]

Remember the constant we talked about in the beginning?

\[=\frac{1}{x}log_{b}e\]

Using the theorems of logarithm³ again, we get

\[= \frac{1}{x \cdot \ln(b)}\]

Note: checkout the logarithmic notation³ to know what \(\ln(b)\) means.

Derivative of Natural Logarithm

\[\frac{d}{dx}[\ln(u)]=\frac{1}{u}\cdot\frac{du}{dx}\]

Derivative of Exponential Functions

\[y = f(x) = b^x\]

Applying \(\ln\) function on both sides

\[\ln y = \ln b^x\]

\[= x \cdot \ln (b)\]

\[\frac{d}{dx}[\ln y]=\frac{d}{dx}[x \ln b]\]

\[\frac{1}{y}\frac{dy}{dx}=\ln b\]

\[\frac{dy}{dx}=y\cdot \ln b\]

\[= b^x \cdot \ln b\]

Generalized Form

\[\frac{d}{dx}[a^{u}]=a^{u}\cdot \ln a\cdot\frac{du}{dx}\]

Inverse Functions

Imagine 2 functions,² \(f(x)\) and \(g(x)\).
If \(f(g(x)) = x\) for every \(x\) in domain of \(g(x)\) and if \(g(f(x)) = x\) for every \(x\) in domain of \(f(x)\), then we say both functions² are inverse of each other.

Example

\[f(x) = 2x\]

and there is also

\[g(x) = \frac{x}{2}\]

Then to check, we try both of our conditions

Condition 1

\[f(g(x)) = 2 \left(\frac{x}{2}\right) = x\]

Condition 2

\[g(f(x)) = \frac{1}{2} \cdot (2x) = x\]

For a function² to have an inverse, one of the important conditions is that it should be a one-to-one function.

\[y = f^{-1}(x) \implies f(y) = x\]

Derivative of Inverse Functions

\[(f^{-1})(x) = \frac{1}{f^{'}(f^{-1}(x))}\]

Therefore,

\[\frac{dy}{dx}=(f^{-1})^{\prime}(x)\Rightarrow\frac{dx}{dy}=f^{\prime}(y)=f^{\prime}(f^{-1}(x))\]