Question

What is the derivative of $y=\ln (\sec (x)+\tan (x))$ ?

Answer 1

Hint: We know that the natural exponential function and natural logarithmic functions are inverses of each other. In this question initially we had to suppose the given function as some variable and differentiate the given function with respect to that variable and at last the obtained result would be the answer.

Complete step-by-step answer:
Now, in the question we are already provided the right-hand side equal to some variable y as $y=\ln (\sec (x)+\tan (x))$ .
Now raising this to the exponential power we get,
${{e}^{y}}=\sec x+\tan x$
Now doing the derivative we get
${{e}^{y}}\dfrac{dy}{dx}=\sec x\tan x+{{\sec }^{2}}x$
Now we can see on the right-hand side $\sec x$ is common in both the terms we can take it out as common and further simplify as
 ${{e}^{y}}\dfrac{dy}{dx}=\sec x(\tan x+\sec x)$
Now on the left-hand side if we keep only the $\dfrac{dy}{dx}$ and divide the right-hand side the exponential term raised to y we get, $\dfrac{dy}{dx}=\dfrac{1}{{{e}^{y}}}\left( \sec x(\sec x+\tan x) \right)$ and from the starting of the solution we know that we can make substitutions know and after doing that we will get $\dfrac{dy}{dx}=\dfrac{1}{(\sec x+\tan x)}\left( \sec x(\sec x+\tan x) \right)$
And now clearly, we can see that the term $\sec x+\tan x$ is present in numerator and denominator both so we can cancel it out and we get, $\dfrac{dy}{dx}=\sec x$ .

Note: All we need to know for this type of questions is the relations between different types of functions and get the required answer easily. Apart from that we must not get confused in the derivative and antiderivative value of logarithmic, exponential and other types of functions which are very much similar.