Question

The derivative of $a^{secx}$ with respect to $a^{\tan x},\left(a>0\right)$ is

• A. $secx{a}^{secx-\tan x}$
• B. $\sin x{a}^{\tan x-secx}$
• C. $\sin x{a}^{secx-\tan x}$
• D. $a^{secx-\tan x}$

Answer 1

The correct option is C

$\sin x{a}^{secx-\tan x}$

Explanation for the correct answer:

Let $u=a^{secx}$ and let $v=a^{\tan x},\left(a>0\right)$

Determine $\frac{du}{dx}$, for $u=a^{secx}$,

Taking log on both sides we have,

$$\begin{gathered} \log u=\log a^{secx} \\ \log u=secx\log a[\because \log a^x=x\log a] \end{gathered}$$

Differentiating with respect to $x$, we have,

$\Rightarrow \frac{1}{u}\frac{du}{dx}=\log asecx\tan x$

$\Rightarrow \frac{du}{dx}=u\log asecx\tan x$

Substitute the value of $u$

$\Rightarrow \frac{du}{dx}=a^{secx}\times \log a\times secx\times \tan x$

Determine $\frac{dv}{dx}$. for $v=a^{\tan x}$

Taking log on both sides we have,

$$\begin{gathered} \log v=\log a^{\tan x} \\ \log v=\tan x\log a\mathbf{[}\mathbf{\because }\mathbf{log}\mathbf{}{\mathbf{a}}^{\mathbf{x}}\mathbf{=}\mathbf{x}\mathbf{}\mathbf{log}\mathbf{}\mathbf{a}\mathbf{]} \end{gathered}$$

Differentiating with respect to $x$, we have,

$\Rightarrow \frac{1}{v}\frac{dv}{dx}=\log a\times se{c}^{2}x$

$\Rightarrow \frac{dv}{dx}=v\log a\times se{c}^{2}x$

Substitute the value of $v$,

$\Rightarrow$$\frac{dv}{dx}=a^{\tan x}\times \log a\times se{c}^{2}x$

We know that, $\frac{du}{dv}=\frac{\frac{du}{dx}}{\frac{dv}{dx}}$

$=\frac{a^{secx}\times \log a\times secx\times \tan x}{a^{\tan x}\times \log a\times se{c}^{2}x}$

$=\left(\frac{\tan x}{secx}\right)\times a^{secx-\tan x}$

$=\left(\frac{\sin x\times \cos x}{\cos x}\right)\times a^{secx-\tan x}\left[\because secx=\frac{1}{cosx}andtanx=\frac{sinx}{cosx}\right]$

$=\sin x\times a^{secx-\tan x}$

Hence, $\frac{du}{dv}=\sin x{a}^{secx-\tan x}$

Therefore, the correct answer is option (C)