In the previous lesson, we have reviewed the concepts and the handling of exponential and logarithmic functions. It is essential to develop a strong understanding of the basic rules and laws governing such functions’ analysis before attempting to try to understand its derivative.
Just as algebraic functions, differentiating exponential and logarithmic functions have its own set of rules. For the following formulas, let $u$ be a function of $x$; $a$ as a constant; and $e$ is the Euler’s number.
$$\begin{align} \text{Logarithmic Funcion to Base $a$:} &&& d(\log_a u) = \frac{du}{u\,\ln a} \label{eq:differential of the Logarithmic Funcion to Base a}\\ \text{Natural Logarithmic Function:} &&& d(\ln u) = \frac{du}{u} \label{eq:differential of Natural Logarithmic Function}\\ \text{Exponential Function, $e^u$:} &&& d(e^u) = e^u\,du \label{eq:differential of Exponential Function e^u}\\ \text{Exponential Function, $a^u$:} &&& d(a^u) = a^u\ln a\,du \label{eq:differential of Exponential Function a^u} \end{align}$$
$\example{1}$ Find the derivative of the function $y = \log_5 (2x+3)$
$\solution$
\[\begin{align*} \text{Let:}\quad a &= 5 \hspace{100cm} \\ u &= 2x+3 & \\ du &= 2\,dx \end{align*}\]
To solve for the derivative of the function, use $\eref{eq:differential of the Logarithmic Funcion to Base a}$,
\[\begin{align*} d(y) &= d\br{\log_5 (2x+3)} \\ dy &= \frac{2\,dx}{(2x+3)\ln 5} \qquad\tcA{d(\log_a u) = \frac{du}{u\,\ln a}} \\ \dydx &= \frac{2}{(2x+3)\ln 5} \tagans \end{align*}\]$\example{2}$ Find the derivative of the function $y = \log_e (5-2x)$
$\solution$
\[\begin{align*} \text{Let:}\quad a &= e \hspace{100cm} \\ u &= 5-2x & \\ du &= -2\,dx \end{align*}\]
To solve for the derivative of the function, use $\eref{eq:differential of the Logarithmic Funcion to Base a}$,
\[\begin{align*} d(y) &= d\br{\log_e (5-2x)} \\ dy &= \frac{-2\,dx}{(5-2x)\ln e} \end{align*}\]But form the Properties of Natural Logarithms, $\ln e = 1$. Thus,
\[\begin{align*} \dydx &= -\frac{2}{(5-2x)(1)} \\ &= -\frac{2}{5-2x} \tagans \end{align*}\]$\example{3}$ Find the derivative of the function $y = \ln (5-2x)$
$\solution$
\[\begin{align*} \text{Let:}\quad u &= 5-2x \hspace{100cm} \\ du &= -2\,dx \end{align*}\]
To solve for the derivative of the function, use $\eref{eq:differential of Natural Logarithmic Function}$,
\[\begin{align*} d(y) &= d\br{\ln (5-2x)} \\ dy &= \frac{-2\,dx}{5-2x} \\ \dydx &= -\frac{2}{5-2x} \tagans \end{align*}\]As you may have noticed, the answer is the same as the previous example. With this, we can verify that $\log_e u$ is equal to $\ln u$.
$\example{4}$ Differentiate the function $y = \ln\sqrt{2x^2-5x+3}$ with respect to $x$.
$\solution$
\[\begin{align*} \text{Let:}\quad u &= \sqrt{2x^2-5x+3} \hspace{100cm} \\ du &= d\br{\sqrt{2x^2-5x+3}} \\ &= d\br{(2x^2-5x+3)^{1/2}} \\ &= \frac12(2x^2-5x+3)^{-1/2}\cdot d(2x^2-5x+3) \\ &= \frac{1}{2\sqrt{2x^2-5x+3}} \cdot (4x-5) \,dx \\ &= \frac{4x-5}{2\sqrt{2x^2-5x+3}} \,dx \\ \end{align*}\]
To solve for the derivative of the function, use $\eref{eq:differential of Natural Logarithmic Function},$
\[\begin{align*} d(y) &= d\br{\ln\sqrt{2x^2-5x+3}} \\ dy &= \frac{\frac{4x-5}{2\sqrt{2x^2-5x+3}} \,dx}{\sqrt{2x^2-5x+3}} \\ dy &= \frac{4x-5}{2\sqrt{2x^2-5x+3} \cdot \sqrt{2x^2-5x+3}} \,dx \\ \dydx &= \frac{4x-5}{2\br{\sqrt{2x^2-5x+3}}^2} \\ &= \frac{4x-5}{2\br{2x^2-5x+3}} \tagans \end{align*}\]$\example{5}$ Determine the differential of the function $y = 5e^{3x}$.
$\solution$
To solve for the differential of the function, use $\eref{eq:differential of Exponential Function e^u}$,
\[\begin{align*} d(y) &= d\br{5e^{3x}} \\ dy &= 5e^{3x} \cdot d(3x) \\ &= 5e^{3x} \cdot 3\,dx \\ &= 15e^{3x} \,dx \tagans \end{align*}\]$\example{6}$ Find the derivative of the function with respect to $t$: $x = e^{3t}e^{2t^2}$.
$\solution$
To solve for the derivative of the function, use $\eref{eq:differential of Exponential Function e^u}$,
\[\begin{align*} d(x) &= d\br{e^{3t}e^{2t^2}} \\ dx &= d\br{e^{3t+2t^2}} \qquad\tcA{a^m+a^n = a^{m+n}} \\ &= e^{3t+2t^2} \cdot d(3t+2t^2) \\ &= e^{3t+2t^2} \cdot (3+4t)\,dt \\ \frac{dx}{dt} &= (3+4t)e^{3t+2t^2} \tagans \end{align*}\]$\example{7}$ Determine the differential of the function $w = 125\br{5^{3v}}$.
$\solution$
To solve for the differential of the function, use $\eref{eq:differential of Exponential Function a^u}$,
\[\begin{align*} d(w) &= d\br{125\br{5^{3v}}} \\ dw &= d\br{5^3\cdot 5^{3v}} \\ &= d\br{5^{3+3v}} \\ &= 5^{3+3v} \ln 5 \cdot d\br{3+3v} \\ &= 5^{3+3v} \ln 5 \cdot 3 \,dv \\ &= 3\ln 5\cdot 5^{3+3v} \,dv \tagans \end{align*}\]$\example{8}$ Find the derivative of $y$ with respect to $x$ of the function $e^{\frac12 y} = x^2-1$.
$\solution$
In this example, the given equation is expressed as an implicit function. Hence, to determine its derivative, we can differentiate the given equation explicitly. \(\begin{align*} d\br{e^{\frac12 y}} &= d\br{x^2-1} \\ e^{\frac12 y} \cdot d\br{\frac12 y} &= 2x\,dx \\ e^{y/2} \cdot \frac12 \,dy &= 2x\,dx \\ \dydx &= \frac{4x}{e^{y/2}} \tagans \end{align*}\)