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# How do you find the derivative of $\sqrt {{x^2} + {y^2}}$ ?

Last updated date: 03rd Mar 2024
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Hint: The above concept is based on derivation of the given expression. The main approach to solve the expression is to reduce the whole power by 1 and then apply chain rule to the terms under the square root to get the derivative of the given expression.

Complete step by step solution:
In mathematics derivative is a way to show instantaneous rate of change. The above given expression has square root sign which needs to be differentiated when the function contains radical sign, the power rule seems difficult to apply. Using a simple exponent substitution, differentiating this function becomes very straightforward and easy.

The above expression can be solved by implicit differentiation where we differentiate each side of an equation with two variables x and y by treating one of the variables as a function of others. This can be done by applying chain rule. The chain rule tells us how to find derivatives of composite function. The chain rule can be written as:
$\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \times \dfrac{{du}}{{dx}}$
Therefore, we can write it has
$\dfrac{d}{{dx}}\left( {\sqrt x } \right) = \dfrac{1}{{2\sqrt x }}$
So now,
$\dfrac{d}{{dx}}\left( {\sqrt u } \right) = \dfrac{1}{{2\sqrt u }}\dfrac{{du}}{{dx}}$
Therefore, now differentiating it,
$\dfrac{d}{{dx}}\left( {\sqrt {{x^2} + {y^2}} } \right) = \dfrac{1}{{2\sqrt {{x^2} + {y^2}} }} \times \dfrac{d}{{dx}}\left( {{x^2} + {y^2}} \right) \\ \Rightarrow\dfrac{d}{{dx}}\left( {\sqrt {{x^2} + {y^2}} } \right) = \dfrac{1}{{2\sqrt {{x^2} + {y^2}} }}\left( {2x + 2y\dfrac{{dy}}{{dx}}} \right) \\$
Now by opening the brackets and multiplying the terms inside it.
$\Rightarrow\dfrac{d}{{dx}}\left( {\sqrt {{x^2} + {y^2}} } \right)= \dfrac{1}{{2\sqrt {{x^2} + {y^2}} }}2x + \dfrac{1}{{2\sqrt {{x^2} + {y^2}} }}2y\dfrac{{dy}}{{dx}} \\ \therefore\dfrac{d}{{dx}}\left( {\sqrt {{x^2} + {y^2}} } \right) = \dfrac{x}{{\sqrt {{x^2} + {y^2}} }} + \dfrac{y}{{\sqrt {{x^2} + {y^2}} }}\dfrac{{dy}}{{dx}}$
Therefore, we get the above solution.

Note: An important thing to note is that the square root is differentiated in such a way that the square power i.e., $\dfrac{1}{2}$ is subtracted by 1 and we get $\dfrac{1}{2} - 1 = - \dfrac{1}{2}$.Since we get negative value in the power the term goes in the denominator and also $\dfrac{1}{2}$ is written by multiplying it.