Question

Differentiate with respective to x:
\[\log \left( {\sec x\,\, + \,\,\tan x} \right)\]

Answer 1

Hint: We will suppose to the given value \[v{\text{ }} = {\text{ }}log{\text{ }}\left( {sec{\text{ }}x{\text{ }} + {\text{ }}tan{\text{ }}x} \right)\]. Further taking log both sides, then differentiate the given value with respect to x.
\[\dfrac{d}{{dx}}\left( {\log x} \right) = \dfrac{1}{x}\]


Complete step by step solution:-
let \[v{\text{ }} = {\text{ }}log{\text{ }}\left( {sec{\text{ }}x{\text{ }} + {\text{ }}tan{\text{ }}x} \right)\]
Differentiate both side with respect to x, we will get
$
  \dfrac{d}{{dx}}v = \dfrac{d}{{dx}}\left( {\log 1\sec + \tan x} \right) \\
  \dfrac{d}{{dx}} = \dfrac{1}{{\sec x + tax}} \times \dfrac{d}{{dx}}\left( {\sec x + \tan x} \right) \\
   = \dfrac{1}{{\left( {\sec x + \tan x} \right)}} \times \sec x.\tan x + {\sec ^2}x \\
  \dfrac{{dv}}{{dx}} = \dfrac{1}{{\sec x + \tan x}}\left( {\sec x - \tan x + {{\sec }^2}x} \right) \\
  \dfrac{{dv}}{{dx}} = \dfrac{1}{{\sec x + \tan x}}\sec x\left( {\tan x + \sec x} \right) \\
  \dfrac{{dv}}{{dx}} = \dfrac{1}{{\left( {\sec x + \tan x} \right)}}\sec x\left( {\sec x + \tan x} \right) \\
  \dfrac{{dv}}{{dx}} = \sec x \\
 $


Additional Information: Differentiation comes down to figuring out how one variable changes with respect to another variable. Some differentiation rule are:
(i) The constant rule: for any fixed real number $c$.\[\dfrac{d}{{dx}}\left\{ {c.f(x)} \right\} = c.\dfrac{d}{{dx}}\left\{ {f(x)} \right\}\]
(ii) The power rule: $\dfrac{d}{{dx}}\left\{ {{x^n}} \right\} = n{x^{n - 1}}$


Note: We have to be careful to use the appropriate formula of logarithm in accordance to the problem given.Some of them are:
 \[\left( i \right)\log {(a)^m} = m\log a\]
\[\left( {ii} \right)\log a.\log b = \log \left( {a + 3} \right)\]
\[
  \left( {iii} \right)\log \left( {\dfrac{a}{b}} \right) = \log a - \log b \\
  \left( {iv} \right)\log 1 = 0 \\
  \left( v \right)\log e = 1 \\
 \]