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Hint: We will be using the concept of differential calculus to solve the problem. We will be using chain rule of differentiation to solve the problem.

__Complete step-by-step answer:__

Now, we have to differentiate $3{{x}^{2}}-{{e}^{-3x}}+\sec x$ with respect to x.

So, we let,

$f\left( x \right)=3{{x}^{2}}-{{e}^{-3x}}+\sec x$

Now, we know that if $f\left( x \right)=g\left( x \right)+h\left( x \right)$ then,

$\dfrac{d}{dx}f\left( x \right)=\dfrac{d}{dx}g\left( x \right)+\dfrac{d}{dx}h\left( x \right)$

Therefore, we have,

$\dfrac{d}{dx}f\left( x \right)=\dfrac{d}{dx}\left( 3{{x}^{2}} \right)-\dfrac{d}{dx}\left( {{e}^{-3x}} \right)+\dfrac{d}{dx}\left( \sec x \right)$

Now, we know that,

$\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$

Now, we know that according to chain rule,

\[\begin{align}

& \dfrac{d}{dx}\left( {{e}^{-ax}} \right)={{e}^{-ax}}\times \dfrac{d}{dx}\left( -ax \right) \\

& \dfrac{d}{dx}\left( \sec x \right)=\sec x\tan x \\

\end{align}\]

So, we have,

\[\begin{align}

& \dfrac{d}{dx}f\left( x \right)=3\times 2x-\left( -3 \right){{e}^{-3x}}+\sec x\tan x \\

& =6x+3{{e}^{-3x}}+\sec x\tan x \\

\end{align}\]

Therefore, we have the differentiation of $3{{x}^{2}}-{{e}^{-3x}}+\sec x$ with respect to x as,

\[6x+3{{e}^{-3x}}+\sec x\tan x\]

Note: To solve these type of questions it is important to remember that if $f\left( x \right)=g\left( x \right)+h\left( x \right)$ then $f'\left( x \right)=g'\left( x \right)+h'\left( x \right)$. Also, it should be noted $\dfrac{d}{dx}\left( {{e}^{-3x}} \right)$ have been found by using chain rule.

Now, we have to differentiate $3{{x}^{2}}-{{e}^{-3x}}+\sec x$ with respect to x.

So, we let,

$f\left( x \right)=3{{x}^{2}}-{{e}^{-3x}}+\sec x$

Now, we know that if $f\left( x \right)=g\left( x \right)+h\left( x \right)$ then,

$\dfrac{d}{dx}f\left( x \right)=\dfrac{d}{dx}g\left( x \right)+\dfrac{d}{dx}h\left( x \right)$

Therefore, we have,

$\dfrac{d}{dx}f\left( x \right)=\dfrac{d}{dx}\left( 3{{x}^{2}} \right)-\dfrac{d}{dx}\left( {{e}^{-3x}} \right)+\dfrac{d}{dx}\left( \sec x \right)$

Now, we know that,

$\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$

Now, we know that according to chain rule,

\[\begin{align}

& \dfrac{d}{dx}\left( {{e}^{-ax}} \right)={{e}^{-ax}}\times \dfrac{d}{dx}\left( -ax \right) \\

& \dfrac{d}{dx}\left( \sec x \right)=\sec x\tan x \\

\end{align}\]

So, we have,

\[\begin{align}

& \dfrac{d}{dx}f\left( x \right)=3\times 2x-\left( -3 \right){{e}^{-3x}}+\sec x\tan x \\

& =6x+3{{e}^{-3x}}+\sec x\tan x \\

\end{align}\]

Therefore, we have the differentiation of $3{{x}^{2}}-{{e}^{-3x}}+\sec x$ with respect to x as,

\[6x+3{{e}^{-3x}}+\sec x\tan x\]

Note: To solve these type of questions it is important to remember that if $f\left( x \right)=g\left( x \right)+h\left( x \right)$ then $f'\left( x \right)=g'\left( x \right)+h'\left( x \right)$. Also, it should be noted $\dfrac{d}{dx}\left( {{e}^{-3x}} \right)$ have been found by using chain rule.

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