Question

If $y=\log \left(x\right)e^{\left(\tan x+x^2\right)}$ , then $\frac{dy}{dx}$ is equal to?

• A. $e^{\left(\tan x+{\left(x\right)}^2\right)}\left(\frac{1}{x}+\left(se{c}^{2}x+x\right)\log \left(x\right)\right)$
• B. $e^{\left(\tan x+{\left(x\right)}^2\right)}\left(\frac{1}{x}+\left(se{c}^{2}x-x\right)\log \left(x\right)\right)$
• C. $e^{\left(\tan x+{\left(x\right)}^2\right)}\left(\frac{1}{x}+\left(se{c}^{2}x+2x\right)\log \left(x\right)\right)$
• D. $e^{\left(\tan x+{\left(x\right)}^2\right)}\left(\frac{1}{x}+\left(se{c}^{2}x-2x\right)\log \left(x\right)\right)$

Answer 1

The correct option is C

$e^{\left(\tan x+{\left(x\right)}^2\right)}\left(\frac{1}{x}+\left(se{c}^{2}x+2x\right)\log \left(x\right)\right)$

Explanation for the correct option:

To find the value of $\frac{dy}{dx}$ :

Given,

$y=\log \left(x\right)e^{\left(\tan x+x^2\right)}$

Now, differentiate the given function with respect to $x$,

Then,

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{1}{x}{e}^{(\tan x+x^2)}+\log \left(x\right)e^{(\tan x+x^2)}\left(se{c}^{2}x+2x\right)\left[\because \frac{d}{dx}\left(u.v\right)=u\frac{dv}{dx}+v\frac{du}{dv},\frac{d\left(\tan x\right)}{dx}=se{c}^{2}x\right]\\ & =& e^{(\tan x+x^2)}+\left(\frac{1}{x}+\left(se{c}^{2}x+2x\right)\log \left(x\right)\right)\end{array}$

Hence, the correct option is C.