Answer
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Hint: In the given question, we need to find the derivative of the given function that involves square root within it. We know that derivatives are the rate of change of one variable with respect to another variable. It is also the slope of the tangent.
Complete step-by-step solution:
According to the given question, we need to find the derivative of the function $y=\sqrt{x}+\dfrac{1}{\sqrt{x}}$. Derivatives are basically known as the rate of change of one variable with respect to another variable.
Now, we know that derivative of ${{x}^{n}}=n{{x}^{n-1}}$ . So, using this in the given question we will get the derivative of the function which is denoted by $y'\left( x \right)$ .
Therefore,
\[\begin{align}
& y=\sqrt{x}+\dfrac{1}{\sqrt{x}} \\
& \Rightarrow y'\left( x \right)=\dfrac{1}{2\sqrt{x}}-\dfrac{1}{2x\sqrt{x}} \\
& \Rightarrow y'\left( x \right)=\dfrac{\left( x-1 \right)}{2x\sqrt{x}} \\
\end{align}\]
Now, in this first term we have n as half and for second term we have n as negative of half and then simply applying the previously mentioned formula used for derivation of polynomials we got the complete derivative of the function.
Here we have to find the value of $\dfrac{dy}{dx}$ at x=1 so we put this value in $y'\left( x \right)=\dfrac{\left( x-1 \right)}{2x\sqrt{x}}$.
Hence,
$\Rightarrow y'\left( x \right)=\dfrac{\left( 1-1 \right)}{2x\sqrt{1}} = 0 $
Hence, the derivative of the given function is \[y'\left( x \right)=\dfrac{\left( x-1 \right)}{2\sqrt{x}}\] and its value at x=1 is 0. So the option D is the correct answer.
Note: In such a type of question, we must be aware of the definition and how to evaluate the derivative. Also, we need to memorize the results or we can say derivatives of all kinds of functions in order to quickly write the result and move further. Some different functions whose derivatives must be memorized are trigonometric, logarithmic, exponential, polynomial etc.
Complete step-by-step solution:
According to the given question, we need to find the derivative of the function $y=\sqrt{x}+\dfrac{1}{\sqrt{x}}$. Derivatives are basically known as the rate of change of one variable with respect to another variable.
Now, we know that derivative of ${{x}^{n}}=n{{x}^{n-1}}$ . So, using this in the given question we will get the derivative of the function which is denoted by $y'\left( x \right)$ .
Therefore,
\[\begin{align}
& y=\sqrt{x}+\dfrac{1}{\sqrt{x}} \\
& \Rightarrow y'\left( x \right)=\dfrac{1}{2\sqrt{x}}-\dfrac{1}{2x\sqrt{x}} \\
& \Rightarrow y'\left( x \right)=\dfrac{\left( x-1 \right)}{2x\sqrt{x}} \\
\end{align}\]
Now, in this first term we have n as half and for second term we have n as negative of half and then simply applying the previously mentioned formula used for derivation of polynomials we got the complete derivative of the function.
Here we have to find the value of $\dfrac{dy}{dx}$ at x=1 so we put this value in $y'\left( x \right)=\dfrac{\left( x-1 \right)}{2x\sqrt{x}}$.
Hence,
$\Rightarrow y'\left( x \right)=\dfrac{\left( 1-1 \right)}{2x\sqrt{1}} = 0 $
Hence, the derivative of the given function is \[y'\left( x \right)=\dfrac{\left( x-1 \right)}{2\sqrt{x}}\] and its value at x=1 is 0. So the option D is the correct answer.
Note: In such a type of question, we must be aware of the definition and how to evaluate the derivative. Also, we need to memorize the results or we can say derivatives of all kinds of functions in order to quickly write the result and move further. Some different functions whose derivatives must be memorized are trigonometric, logarithmic, exponential, polynomial etc.
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