Question

How do you find the first and second derivative of $y = \dfrac{1}{{1 + {e^x}}}$ ?

Answer 1

Hint : We know that functions with base e, is often referred to as the exponential function. Exponential functions and their corresponding inverse functions, are logarithmic functions i.e., The natural logarithm, or logarithm to base e, is the inverse function to the natural exponential function. To find the first derivative of the given function, differentiate the function with respect to x and then to find the second derivative of the given function, we need to apply quotient rule to the first derivative.

Complete step by step solution:
Given,
$y = \dfrac{1}{{1 + {e^x}}}$
The given function is re-written as:
$y = {\left( {1 + {e^x}} \right)^{ - 1}}$ ……………………. 1
To find the first derivative, we need to apply chain rule to the denominator term i.e., $1 + {e^x}$ , in which the chain rule is defined as: If, $f = v\left( {u\left( x \right)} \right)$ , then, the derivative of function f is:
$\Rightarrow \dfrac{{df}}{{dx}} = \dfrac{{dv}}{{dt}} \times \dfrac{{dt}}{{dx}}$
Differentiating equation 1 with respect to x as follows:
$\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}{\left( {1 + {e^x}} \right)^{ - 1}}$
We get:
$\Rightarrow \dfrac{{dy}}{{dx}} = - {\left( {1 + {e^x}} \right)^{ - 2}}\dfrac{d}{{dx}}\left( {1 + {e^x}} \right)$
We know that, $\dfrac{d}{{dx}}\left( {1 + {e^x}} \right) = {e^x}$ , hence we get:
$\Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{1}{{{{\left( {1 + {e^x}} \right)}^2}}}{e^x}$
$\Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{{{e^x}}}{{{{\left( {1 + {e^x}} \right)}^2}}}$ …………………… 2
Hence, the first derivative of $y = \dfrac{1}{{1 + {e^x}}}$ is:
$\dfrac{{dy}}{{dx}}\left( {\dfrac{1}{{1 + {e^x}}}} \right) = - \dfrac{{{e^x}}}{{{{\left( {1 + {e^x}} \right)}^2}}}$ .
Now, to find second derivative we need to apply quotient rule which states:
$\dfrac{d}{{dx}}\left( {\dfrac{{u\left( x \right)}}{{v\left( x \right)}}} \right) = \dfrac{{u'\left( x \right) \cdot v\left( x \right) - u\left( x \right) \cdot v'\left( x \right)}}{{v{{\left( x \right)}^2}}}$
Here, from equation 2 we have:
$u\left( x \right) = {e^x}$
$\Rightarrow u'\left( x \right) = {e^x}$
$v\left( x \right) = {\left( {1 + {e^x}} \right)^2}$
$\Rightarrow v'\left( x \right) = 2 \cdot \left( {1 + {e^x}} \right) \cdot {e^x}$
Now, let us substitute these values in the quotient rule as:
$\dfrac{d}{{dx}}\left( {\dfrac{{u\left( x \right)}}{{v\left( x \right)}}} \right) = \dfrac{{u'\left( x \right) \cdot v\left( x \right) - u\left( x \right) \cdot v'\left( x \right)}}{{v{{\left( x \right)}^2}}}$
$\Rightarrow \dfrac{{{d^2}}}{{d{x^2}}}\left( {\dfrac{1}{1} + {e^x}} \right) = \dfrac{{{e^x} \cdot {{\left( {1 + {e^x}} \right)}^2} - {e^x} \cdot {e^x} \cdot 2 \cdot \left( {1 + {e^x}} \right)}}{{{{\left( {{{\left( {1 + {e^x}} \right)}^2}} \right)}^2}}}$
Simplify and combine the numerator terms, we get:
$= \dfrac{{{e^x} \cdot \left( {1 + {e^x}} \right)\left( {\left( {1 + {e^x}} \right) - 2{e^x}} \right)}}{{{{\left( {1 + {e^x}} \right)}^4}}}$
Now, the obtained equation consists of like terms i.e., $\left( {1 + {e^x}} \right)$ ; hence evaluating the numerator and denominator terms, we get:
$= \dfrac{{{e^x} \cdot \left( {\left( {1 + {e^x}} \right) - 2{e^x}} \right)}}{{{{\left( {1 + {e^x}} \right)}^3}}}$
Evaluating the numerator terms, we get:
$= \dfrac{{{e^x} \cdot \left( {1 + {e^x} - 2{e^x}} \right)}}{{{{\left( {1 + {e^x}} \right)}^3}}}$
$= \dfrac{{{e^x} \cdot \left( {1 - {e^x}} \right)}}{{{{\left( {1 + {e^x}} \right)}^3}}}$
Therefore, the second derivative of $y = \dfrac{1}{{1 + {e^x}}}$ is:
$\dfrac{{{d^2}y}}{{d{x^2}}}\left( {\dfrac{1}{{1 + {e^x}}}} \right) = \dfrac{{{e^x} \cdot \left( {1 - {e^x}} \right)}}{{{{\left( {1 + {e^x}} \right)}^3}}}$ .
So, the correct answer is “ $\dfrac{{{d^2}y}}{{d{x^2}}}\left( {\dfrac{1}{{1 + {e^x}}}} \right) = \dfrac{{{e^x} \cdot \left( {1 - {e^x}} \right)}}{{{{\left( {1 + {e^x}} \right)}^3}}}$”.

Note : Note that the exponential function $f\left( x \right) = {e^x}$ has the special property that its derivative is the function itself, $f'\left( x \right) = {e^x} = f\left( x \right)$ . We must know that, to multiply powers with the same base, add the exponents and keep the common base, to divide powers with the same base, subtract the exponents and keep the common base. To raise a power to a power, keep the base and multiply the exponents and to raise a quotient to a power, raise the numerator and the denominator to the power.