Question

How to solve this derivative? With steps $ \dfrac{1}{{{e^x} + \sqrt {{e^x} + 1} }} $

Answer 1

Hint: We will use the derivative of the inverse function formula to find the derivative in the first step then later we will also use the chain rule of the composite function. Then we will apply the derivative of the given power in the function and then by using the chain rule will take the derivative of the function and then the derivative of the angle. In between also use the law of negative power and its equivalent positive term.

Complete step by step solution:
Take the given expression: $ f(x) = \dfrac{1}{{{e^x} + \sqrt {{e^x} + 1} }} $
Apply the reciprocal rule in the above expression,
 $ f'(x) = - \dfrac{{\dfrac{d}{{dx}}[{e^x} + \sqrt {{e^x} + 1} ]}}{{{{({e^x} + \sqrt {{e^x} + 1} )}^2}}} $
Apply derivative to both the terms in the numerator.
 $ f'(x) = - \dfrac{{\dfrac{d}{{dx}}[{e^x}] + \dfrac{d}{{dx}}(\sqrt {{e^x} + 1} )]}}{{{{({e^x} + \sqrt {{e^x} + 1} )}^2}}} $
Place the formula for the derivatives in the above equation, Also apply chain rule where applicable.
 $ f'(x) = - \dfrac{{[{e^x}] + \dfrac{1}{2}{{({e^x} + 1)}^{\dfrac{1}{2} - 1}}.\dfrac{d}{{dx}}[{e^x} + 1]}}{{{{({e^x} + \sqrt {{e^x} + 1} )}^2}}} $
Simplify the above equation-
 $ f'(x) = - \dfrac{{[{e^x}] + \dfrac{1}{2}{{({e^x} + 1)}^{ - \dfrac{1}{2}}}.[{e^x} + 0]}}{{{{({e^x} + \sqrt {{e^x} + 1} )}^2}}} $
The negative one by two power is equal to the positive square root in the denominator.
 $ f'(x) = - \dfrac{{[{e^x}] + \dfrac{1}{2}\dfrac{{{e^x}}}{{\sqrt {({e^x} + 1)} }}}}{{{{({e^x} + \sqrt {{e^x} + 1} )}^2}}} $
This is the required solution.
So, the correct answer is “ $ \dfrac{{[{e^x}] + \dfrac{1}{2}\dfrac{{{e^x}}}{{\sqrt {({e^x} + 1)} }}}}{{{{({e^x} + \sqrt {{e^x} + 1} )}^2}}} $ ”.

Note: Know the difference between the differentiation and the integration and apply formula accordingly. Differentiation can be represented as the rate of change of the function, whereas integration represents the sum of the function over the range. They are inverses of each other.