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**Hint:**We will solve this question by chain rule , chain rule states that derivative of f(g(x)) with respect to x is equal to derivative of f(g(x)) with respect to g multiplied by derivative of g(x) with respect to x. we know that derivative of log x is $\dfrac{1}{x}$

**Complete step by step solution:**

We have to evaluate $\dfrac{d}{dx}\left\{ \log \left( x+\sqrt{{{a}^{2}}+{{x}^{2}}} \right) \right\}$

Derivative of $\log \left( x+\sqrt{{{a}^{2}}+{{x}^{2}}} \right)$ with respect to x let us take $x+\sqrt{{{a}^{2}}+{{x}^{2}}}$ equal to t

So we can write $\dfrac{d}{dx}\left\{ \log t \right\}=\dfrac{d}{dt}\left\{ \log t \right\}\times \dfrac{dt}{dx}$

We know that derivative of log t with respect to t is equal to $\dfrac{1}{t}$

We can write $\dfrac{d}{dx}\left\{ \log t \right\}=\dfrac{1}{t}\dfrac{dt}{dx}=\dfrac{1}{x+\sqrt{{{a}^{2}}+{{x}^{2}}}}\dfrac{d}{dx}\left( x+\sqrt{{{a}^{2}}+{{x}^{2}}} \right)$

Now we have to find the derivative of $x+\sqrt{{{a}^{2}}+{{x}^{2}}}$ with respect to x , we know that derivative of x with respect to x is equal to 1 and let’s find the derivative of $\sqrt{{{a}^{2}}+{{x}^{2}}}$ with respect to x , if we take ${{a}^{2}}+{{x}^{2}}$ equal to t

We can write $\dfrac{d\sqrt{t}}{dx}=\dfrac{d\sqrt{t}}{dt}\times \dfrac{dt}{dx}$

$\Rightarrow \dfrac{d\sqrt{t}}{dx}=\dfrac{1}{2\sqrt{t}}\times \dfrac{dt}{dx}$

$\Rightarrow \dfrac{d\sqrt{t}}{dx}=\dfrac{1}{2\sqrt{{{a}^{2}}+{{x}^{2}}}}\times 2x$

So the derivative of $\sqrt{{{a}^{2}}+{{x}^{2}}}$ with respect to x is $\dfrac{x}{\sqrt{{{a}^{2}}+{{x}^{2}}}}$

So derivative of $x+\sqrt{{{a}^{2}}+{{x}^{2}}}$ with respect to x is $1+\dfrac{x}{\sqrt{{{a}^{2}}+{{x}^{2}}}}=\dfrac{x+\sqrt{{{a}^{2}}+{{x}^{2}}}}{\sqrt{{{a}^{2}}+{{x}^{2}}}}$

We already know $\dfrac{d}{dx}\left\{ \log \left( x+\sqrt{{{a}^{2}}+{{x}^{2}}} \right) \right\}$ is equal to $\dfrac{1}{x+\sqrt{{{a}^{2}}+{{x}^{2}}}}\dfrac{d}{dx}\left( x+\sqrt{{{a}^{2}}+{{x}^{2}}} \right)$

$\Rightarrow \dfrac{d}{dx}\left\{ \log \left( x+\sqrt{{{a}^{2}}+{{x}^{2}}} \right) \right\}=\dfrac{1}{x+\sqrt{{{a}^{2}}+{{x}^{2}}}}\dfrac{x+\sqrt{{{a}^{2}}+{{x}^{2}}}}{\sqrt{{{a}^{2}}+{{x}^{2}}}}=\dfrac{1}{\sqrt{{{a}^{2}}+{{x}^{2}}}}$

**So, the correct answer is “Option D”.**

**Note:**Derivative of f( x ) with respect to x at a is denoted by f ‘( a) is the value of slope of the tangent drawn at point ( a, f(a) ) in the curve of f. if the derivative value is 0 then the tangent will be parallel to X axis and if the derivative tends to infinity the tangent will be parallel to Y axis.

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