How do you simplify \[\sqrt{170}\]?

Hint: Using the distributive law and the Binomial theorem. Since we have to find square root so first of all assume a function \[y=f(x)=\sqrt{x}\] where \[x\] is number whose square root we have known means easily calculable and also, we know the square root of one of its adjacent number. Then the difference in number would be \[\Delta x\] and the number whose square root has to be calculated will be \[x+\Delta x\]. Now in the function of \[f(x)\] we use the differentiation, now differentiate the function and it would be written as \[\Delta y/\Delta x\] and this will be equal to derivative of assumed function and we know that \[\Delta y=f(x+\Delta x)-f(x)\], function \[f\] is square root (\[sqrt\]). Further on simplifying we will get the required square root as our answer.

Complete step by step solution:
Since we need to calculate the square root of \[\sqrt{170}\],
When we know the \[sqrt\] of a number adjacent to the required number then we use method of differentiation.
Let \[y=f(x)=\sqrt{x}\]
\[\dfrac{dy}{dx}=\dfrac{d}{dx}\left( \sqrt{x} \right)\]
\[\dfrac{dy}{dx}=\dfrac{1}{2\sqrt{x}}\]
Since we know the \[sqrt\] of \[169\]
\[\Rightarrow\] Let \[x=169\]
Now the difference is
\[\Rightarrow \Delta x=1\]
\[y=f(x+\Delta x)=\sqrt{170}\]
\[f(x)=\sqrt{169}\]
\[\Rightarrow \Delta y=f(x+\Delta x)-f(x)\]
\[\Rightarrow { }=\sqrt{170}-13\]
Also, we know that
\[\begin{align}
  & \dfrac{\Delta y}{\Delta x}=\dfrac{dy}{dx} \\
 & \Rightarrow \Delta y=\dfrac{dy}{dx}.\Delta x \\
\end{align}\]
Now substituting the value of \[\Delta y\] in above equation
\[\Rightarrow \sqrt{170}-13=\dfrac{1}{2\sqrt{x}}.\Delta x\]
\[\Rightarrow \sqrt{170}=\dfrac{1}{2\sqrt{x}}.\Delta x+13\]
Now putting the values
\[x=169,\Delta x=1\]
\[\Rightarrow \sqrt{170}=13+\dfrac{1}{2\sqrt{169}}.1\]
\[\begin{align}
  & =13+\dfrac{1}{26} \\
 & =13+0.0384 \\
 & =13.0384 \\
\end{align}\]
Thus, we have calculated the \[\sqrt(170)\] by the method of differentiation.

Hence, \[\sqrt{170}=13.0384\]

Note: For finding the square root of this type of number means when we know the square root of the adjacent number of given number.