Question

How do I find the derivative of $y = \ln ({e^{ - x}} + x{e^{ - x}})$ ?

Answer 1

Hint: In this question, we need to differentiate the given function with respect to x. Firstly, we will factor out the common term inside the parenthesis. Then we will apply one of the properties of logarithms which is given by $\ln (a \cdot b) = \ln a + \ln b$. Also we use the fact that $\ln ({e^a}) = a$. Then the obtained expression we will differentiate using the sum rule of differentiation. After that we will simplify it and obtain the required derivative.

Complete step by step solution:
Given an expression of the form $y = \ln ({e^{ - x}} + x{e^{ - x}})$ …… (1)
We are asked to find the derivative of the above expression.
The given function is a logarithmic function. The logarithmic function is represented as ${\log _b}a$, where b is called the base and a is a number.
Here we have given the natural logarithmic function where it’s base is $e$ and it is represented as $\ln $.
Firstly, we will factor out the common term inside the parenthesis in the equation (1).
So factoring out ${e^{ - x}}$ within the parenthesis, we get,
$y = \ln ({e^{ - x}}(1 + x))$
Now we apply the property of logarithm which is given by, $\ln (a \cdot b) = \ln a + \ln b$
Here $a = {e^{ - x}}$ and $b = 1 + x$
Hence the above expression becomes,
$y = \ln ({e^{ - x}}) + \ln (1 + x)$
Now for the first term, we use the fact that $\ln ({e^a}) = a$.
Here we have $a = - x$. So applying this we get,
$y = - x + \ln (1 + x)$
Now we will differentiate the expression obtained with respect to x.
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}( - x + \ln (1 + x)$
We will use the sum rule of differentiation here, which is given by,
$\dfrac{d}{{dx}}(u + v) = \dfrac{d}{{dx}}u + \dfrac{d}{{dx}}v$
Here $u = - x$ and $v = \ln (1 + x)$
Hence using sum rule, we get,
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}( - x) + \dfrac{d}{{dx}}\ln (1 + x)$
In the first term taking out the constant term which is -1, we get,
$ \Rightarrow \dfrac{{dy}}{{dx}} = ( - 1)\dfrac{{dx}}{{dx}} + \dfrac{d}{{dx}}\ln (1 + x)$
$ \Rightarrow \dfrac{{dy}}{{dx}} = ( - 1)(1) + \dfrac{d}{{dx}}\ln (1 + x)$
We know that $\dfrac{d}{{dx}}\ln x = \dfrac{1}{x}$
So we get,
$ \Rightarrow \dfrac{{dy}}{{dx}} = - 1 + \dfrac{1}{{1 + x}}$
Now we add the fractions, we get,
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - (1 + x) + 1}}{{1 + x}}$
Simplifying we get,
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - x - 1 + 1}}{{1 + x}}$
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - x + 0}}{{1 + x}}$
$ \Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{x}{{1 + x}}$

Hence the derivative of the function $y = \ln ({e^{ - x}} + x{e^{ - x}})$ is given by $\dfrac{{dy}}{{dx}} = - \dfrac{x}{{1 + x}}$.

Note: If the question has the word log or $\ln $, it represents the given function as logarithmic function. Note that we have two types of logarithmic function.
One is a common logarithmic function which is represented as a log and its base is 10.
The other one is a natural logarithmic function represented as $\ln $and its base is $e$.
We must know the basic properties of logarithmic functions and note that these properties hold for both log and $\ln $ functions.
Here we must be careful while applying the property related to logarithmic function.
We must know the derivative of the logarithmic function given by $\dfrac{d}{{dx}}\ln x = \dfrac{1}{x}$.
Some properties of logarithmic functions are given below.
(1) $\ln (x \cdot y) = \ln x + \ln y$
(2) $\ln \left( {\dfrac{x}{y}} \right) = \ln x - \ln y$
(3) $\ln {x^n} = n\ln x$
(4) $\ln 1 = 0$
(5) ${\log _e}e = 1$