Question

The derivative of ${a^{\sec x}}$ with respect to ${a^{\tan x}}$ where $a > 0$ is;
$\left( 1 \right)\sec x{a^{\sec x - \tan x}}$
$\left( 2 \right)\sin x{a^{\tan x - \sec x}}$
$\left( 3 \right)\sin x{a^{\sec x - \tan x}}$
$\left( 4 \right){a^{\sec x - \tan x}}$

Answer 1

Hint: We should be familiar with basic trigonometric identities and operation in order to solve this question. This question requires the knowledge of the concept of differentiation of one function with respect to another function. Let the two functions are $f\left( x \right){\text{ and g}}\left( x \right)$ respectively and according to the concept we have to calculate the value of $\dfrac{{df\left( x \right)}}{{dg\left( x \right)}}$ , which can also be written as $\left[ {\dfrac{{\dfrac{{df\left( x \right)}}{{dx}}}}{{\dfrac{{dg\left( x \right)}}{{dx}}}}} \right]$ means ultimately we have to calculate $\dfrac{{f'\left( x \right)}}{{g'\left( x \right)}}$ .

Complete step-by-step solution:
Since the two given functions have to be differentiated individually, therefore;
Let $u = {a^{\sec x}}$ and $v = {a^{\tan x}}$
First , let us calculate the value of $\dfrac{{du}}{{dx}}{\text{ and }}\dfrac{{dv}}{{dx}}$;

Let us first find out the differentiation of $u$ with respect to $x$ ;
$\left( {\because \dfrac{d}{{dx}}{a^x} = {a^x}{{\log }_e}a} \right)$
And also $\left( {\because \dfrac{d}{{dx}}\sec x = \sec x\tan x} \right)$
Using the above formulae and applying the chain rule of differentiation, we get;
$ \Rightarrow \dfrac{{du}}{{dx}} = {a^{\sec x}}{\log _e}a\sec x\tan x{\text{ }}......\left( 1 \right)$
Let us now find out $\dfrac{{dv}}{{dx}}$ by differentiating $v$ with respect to $x$ ;
$\left( {\because \dfrac{d}{{dx}}\tan x = {{\sec }^2}x} \right)$
$ \Rightarrow \dfrac{{dv}}{{dx}} = {a^{\tan x}}{\log _e}a{\sec ^2}x{\text{ }}......\left( 2 \right)$
Now according to the question we have to differentiate $u = {a^{\sec x}}$ with respect to ${a^{\tan x}}$ , means we have to find out $\dfrac{{du}}{{dv}}$ ; therefore dividing equation $\left( 1 \right)$ by equation $\left( 2 \right)$, we get;
$ \Rightarrow \dfrac{{\dfrac{{du}}{{dx}}}}{{\dfrac{{dv}}{{dx}}}} = \dfrac{{{a^{\sec x}}{{\log }_e}a\sec x\tan x{\text{ }}}}{{{a^{\tan x}}{{\log }_e}a{{\sec }^2}x}}$
On further simplifying the above equation, we get;
$ \Rightarrow \dfrac{{du}}{{dv}} = \dfrac{{{a^{\sec x}}\tan x}}{{{a^{\tan x}}\sec x}}$
$\left( {\because \tan x = \dfrac{{\sin x}}{{\cos x}}{\text{ and }}\because \sec x = \dfrac{1}{{\cos x}}} \right)$
Therefore, the above equation can be further simplified as;
$ \Rightarrow \dfrac{{du}}{{dv}} = \dfrac{{{a^{\sec x}} \times \dfrac{{\sin x}}{{\cos x}}}}{{{a^{\tan x}} \times \dfrac{1}{{\cos x}}}}$
Cancellation of $\cos x$ in the numerator and denominator of R.H.S. ;
$ \Rightarrow \dfrac{{du}}{{dv}} = \dfrac{{{a^{\sec x}} \times \sin x}}{{{a^{\tan x}}}}$
$ \Rightarrow \dfrac{{du}}{{dv}} = \sin x{a^{\sec x}}{a^{ - \tan x}}$
The above equation can also be written as;
$ \Rightarrow \dfrac{{du}}{{dv}} = \sin x{a^{\sec x - \tan x}}$
Therefore the differentiation of ${a^{\sec x}}$ with respect to ${a^{\tan x}}$ is $\sin x{a^{\sec x - \tan x}}$ .
Hence the correct answer for this question is option $\left( 3 \right)$ .

Note: The knowledge of standard formulae for derivatives of exponential functions makes the problem solving easier in questions like this. Listed here are some standard derivative formulae: $\left( 1 \right)$ For exponential functions: $\left( a \right)\dfrac{d}{{dx}}{e^x} = {e^x}$ $\left( b \right)\dfrac{d}{{dx}}{a^x} = \ln \left( a \right){a^x}$ . $\left( 2 \right)$ For logarithmic functions: $\left( a \right)\dfrac{d}{{dx}}\ln \left( x \right) = \dfrac{1}{x}$ , $\left( b \right)\dfrac{d}{{dx}}{\log _a}x = \dfrac{1}{{\left( {x\ln \left( a \right)} \right)}}$ . The function ${e^x}$ has a special property, it’s derivative is the function itself.