Question

Differentiate $\log \left( {\sec x + \tan x} \right)$w.r.t x

Answer 1

Hint: We can equate the expression to a variable. Then we can differentiate using the derivatives of basic functions and applying the chain rule of differentiation.

Complete step by step answer:

We have $y = \log \left( {\sec x + \tan x} \right)$. Its derivative w.r.t x is given by,
\[\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {\log \left( {\sec x + \tan x} \right)} \right)\]
We know that derivative of ${\text{logx}}$ is $\dfrac{1}{x}$, and by applying chain rule of differentiation,$\dfrac{d}{{dx}}\left( {f\left( {g\left( x \right)} \right)} \right) = f'\left( {g\left( x \right)} \right) \times g'\left( x \right)$, we get,
\[\dfrac{{dy}}{{dx}} = \dfrac{1}{{\left( {\sec x + \tan x} \right)}} \times \dfrac{d}{{dx}}\left( {\sec x + \tan x} \right)\]
Now we can open the bracket and differentiate,
\[\dfrac{{dy}}{{dx}} = \dfrac{1}{{\left( {\sec x + \tan x} \right)}} \times \left( {\dfrac{d}{{dx}}\sec x + \dfrac{d}{{dx}}\tan x} \right)\]
We know that \[\dfrac{d}{{dx}}\sec x = {\text{ }}\sec x\tan x\] and \[\dfrac{d}{{dx}}\tan x = {\sec ^2}x\], by applying this, we get,
\[\dfrac{{dy}}{{dx}} = \dfrac{1}{{\left( {\sec x + \tan x} \right)}} \times \left( {\tan x\sec x + {{\sec }^2}x} \right)\]
Now we can take ${\text{secx}}$ outside the bracket
 \[\dfrac{{dy}}{{dx}} = \dfrac{{\sec x\left( {\tan x + \sec x} \right)}}{{\left( {\sec x + \tan x} \right)}}\]
After cancelling the common terms, we get,
\[\dfrac{{dy}}{{dx}} = \sec x\]
Therefore, derivative of ${\text{log}}\left( {{\text{secx + tanx}}} \right)$w.r.t x is \[{\text{secx}}\]

Note: The main concept used in this problem is the chain rule of differentiation. According to the chain rule of differentiation, $\dfrac{d}{{dx}}\left( {f\left( {g\left( x \right)} \right)} \right) = f'\left( {g\left( x \right)} \right) \times g'\left( x \right)$. We must know the derivatives of basic functions. Derivatives of \[{\text{secx}}\] and \[{\text{tanx}}\]can be derived by writing them in terms ${\text{sinx}}$ and ${\text{cosx}}$. And then by taking its derivatives using quotient rule of differentiation. We also use simple algebra to simplify the expression that we will get after differentiation. As integration and differentiation are reverse operations , from this problem we can say that $\log \left( {\sec x + \tan x} \right) + C$ is the integral of \[{\text{secx}}\].