Question

Evaluate the following
\dfrac{d}{dx}\left\\{ \log \left( x+\sqrt{{{a}^{2}}+{{x}^{2}}} \right) \right\\}
A. $\dfrac{1}{x+\sqrt{{{a}^{2}}+{{x}^{2}}}}$
B. $\dfrac{x}{\sqrt{{{x}^{2}}+{{a}^{2}}}}$
C. $\dfrac{x}{x+\sqrt{{{a}^{2}}+{{x}^{2}}}}$
D. $\dfrac{1}{\sqrt{{{a}^{2}}+{{x}^{2}}}}$

Answer 1

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.