Question

Find the derivative and the double derivative of the given function $y = \ln x + {e^x}$.
A.$\dfrac{1}{x} - {e^x}\& \dfrac{1}{{{x^2}}} - {e^x}$
B.$\dfrac{1}{x} + {e^x}\& \dfrac{1}{{{x^2}}} - {e^x}$
C.$\dfrac{1}{x} + {e^x}\& \- \dfrac{1}{{{x^2}}} - {e^x}$
D.$\dfrac{1}{x} + {e^x}\& \- \dfrac{1}{{{x^2}}} + {e^x}$

Answer 1

Hint : In this question we have to find the derivative of the two terms which are in the sum here we will find the derivative of the two terms of the function together and then we will again differentiate the obtained derivative of the function.
First derivative $y' = \dfrac{{dy}}{{dx}}$
Second derivative $y'' = \dfrac{{{d^2}y}}{{d{x^2}}}$

Complete step-by-step answer :
Given the function
$y = \ln x + {e^x}$
The function has two terms in hence we can write this as

$$\Rightarrow y' = \dfrac{{dy}}{{dx}} \\\ = \dfrac{{d\left( {\ln x + {e^x}} \right)}}{{dx}} \\\ = \dfrac{{d\left( {\ln x} \right)}}{{dx}} + \dfrac{{d\left( {{e^x}} \right)}}{{dx}} \\\$$

We know
$\Rightarrow \dfrac{{d\left( {\ln x} \right)}}{{dx}} = \dfrac{1}{x}$
$\Rightarrow \dfrac{{d\left( {{e^x}} \right)}}{{dx}} = 1.{e^x} = {e^x}$
So the derivative of this function will be

$$\Rightarrow y' = \dfrac{{d\left( {\ln x} \right)}}{{dx}} + \dfrac{{d\left( {{e^x}} \right)}}{{dx}} \\\ = \dfrac{1}{x} + {e^x} \\\$$

Now we need to again differentiate the above function since we to find the double derivative of the given function whose first derivative is $y' = \dfrac{1}{x} + {e^x}$
We can write this function for its second derivative as
$\Rightarrow y'' = \dfrac{{dy'}}{{dx}} = \dfrac{{d\left( {\dfrac{1}{x} + {e^x}} \right)}}{{dx}}$
Now since we know the differentiation of an inverse function x is written as
$\Rightarrow \dfrac{{d\left( {\dfrac{1}{x}} \right)}}{{dx}} = \dfrac{{d\left( {{x^{ - 1}}} \right)}}{{dx}} = - {x^{ - 1 - 1}} = - {x^{ - 2}} = - \dfrac{1}{{{x^2}}}$
Hence the differentiation of the above function $y'$will be equal to

$$\Rightarrow y'' = \dfrac{{dy'}}{{dx}} \\\ = \dfrac{{d\left( {\dfrac{1}{x}} \right)}}{{dx}} + \dfrac{{d\left( {{e^x}} \right)}}{{dx}} \\\ = - \dfrac{1}{{{x^2}}} + {e^x}\left[ {\because \dfrac{{d\left( {{e^x}} \right)}}{{dx}} = {e^x}} \right] \\\$$

Hence the second differentiation of the given function $y = \ln x + {e^x}$as$- \dfrac{1}{{{x^2}}} + {e^x}$
Hence the first derivative of the function is $y' = \dfrac{1}{x} + {e^x}$and second derivative is $y'' = - \dfrac{1}{{{x^2}}} + {e^x}$
So, the correct answer is “Option D”.

Note : The derivative of a function y=f(x) of whose variable is x is the measure of rate at which the value y of the function changes with respect to the change of the variable x. Students must note that the second derivative is used to find the nature of the function x whether they are maxima or minima and if the second derivative of the function is zero then the point is the inflection point.