Question

What is the derivative of $w=\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}?$

Answer 1

Hint: We solve this question by using the concept of partial differentiation since the question involves a multivariable function. We partially differentiate the function w with respect to x, y and z individually. Here, x, y and z are independent variables.

Complete step by step solution:
In order to solve this question, let us consider the equation given in the question.
$\Rightarrow w=\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}$
Now, let us first differentiate the function with respect to the variable x partially. This means that we consider the variables y and z as constants while partially differentiating with respect to x.
$\Rightarrow \dfrac{\partial w}{\partial x}=\dfrac{\partial }{\partial x}\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}$
The square root can be written as the power $\dfrac{1}{2}.$
$\Rightarrow \dfrac{\partial w}{\partial x}=\dfrac{\partial }{\partial x}{{\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)}^{\dfrac{1}{2}}}$
We use the chain rule to differentiate this function. By the use of the chain rule, we differentiate the outer function first followed by the inner function. The outer function in this case is ${{\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)}^{\dfrac{1}{2}}}$ and the inner function is ${{x}^{2}}+{{y}^{2}}+{{z}^{2}}.$ Differentiating the outer function first,
$\Rightarrow \dfrac{\partial w}{\partial x}=\dfrac{1}{2}{{\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)}^{\dfrac{1}{2}-1}}.\dfrac{\partial }{\partial x}\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)$
Here, the ${{y}^{2}}$ and ${{z}^{2}}$ terms are treated as constants. Differentiating the inner function,
$\Rightarrow \dfrac{\partial w}{\partial x}=\dfrac{1}{2}{{\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)}^{-\dfrac{1}{2}}}.2x$
Cancelling the 2 in the numerator and denominator and since ${{\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)}^{-\dfrac{1}{2}}}$ can be written as $\dfrac{1}{\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}}.$
$\Rightarrow \dfrac{\partial w}{\partial x}=\dfrac{x}{\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}}$
Similarly, we differentiate with respect to y and z. Partially differentiating with respect to y,
$\Rightarrow \dfrac{\partial w}{\partial y}=\dfrac{\partial }{\partial y}\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}$
The square root can be written as the power $\dfrac{1}{2}.$
$\Rightarrow \dfrac{\partial w}{\partial y}=\dfrac{\partial }{\partial y}{{\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)}^{\dfrac{1}{2}}}$
By the use of the chain rule, we differentiate the outer function first followed by the inner function. The outer function in this case is ${{\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)}^{\dfrac{1}{2}}}$ and the inner function is ${{x}^{2}}+{{y}^{2}}+{{z}^{2}}.$ Differentiating the outer function first,
$\Rightarrow \dfrac{\partial w}{\partial y}=\dfrac{1}{2}{{\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)}^{\dfrac{1}{2}-1}}.\dfrac{\partial }{\partial y}\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)$
Here, the ${{x}^{2}}$ and ${{z}^{2}}$ terms are treated as constants. Differentiating the inner function,
$\Rightarrow \dfrac{\partial w}{\partial y}=\dfrac{1}{2}{{\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)}^{-\dfrac{1}{2}}}.2y$
Cancelling the 2 in the numerator and denominator and since ${{\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)}^{-\dfrac{1}{2}}}$ can be written as $\dfrac{1}{\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}}.$
$\Rightarrow \dfrac{\partial w}{\partial y}=\dfrac{y}{\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}}$
Similarly, now we differentiate with respect to z. Partially differentiating the given function with respect to z,
$\Rightarrow \dfrac{\partial w}{\partial z}=\dfrac{\partial }{\partial z}\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}$
The square root can be written as the power $\dfrac{1}{2}.$
$\Rightarrow \dfrac{\partial w}{\partial z}=\dfrac{\partial }{\partial z}{{\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)}^{\dfrac{1}{2}}}$
By the use of the chain rule, we differentiate the outer function first followed by the inner function. The outer function in this case is ${{\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)}^{\dfrac{1}{2}}}$ and the inner function is ${{x}^{2}}+{{y}^{2}}+{{z}^{2}}.$ Differentiating the outer function first,
$\Rightarrow \dfrac{\partial w}{\partial z}=\dfrac{1}{2}{{\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)}^{\dfrac{1}{2}-1}}.\dfrac{\partial }{\partial z}\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)$
Here, the ${{x}^{2}}$ and ${{y}^{2}}$ terms are treated as constants. Differentiating the inner function,
$\Rightarrow \dfrac{\partial w}{\partial z}=\dfrac{1}{2}{{\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)}^{-\dfrac{1}{2}}}.2y$
Cancelling the 2 in the numerator and denominator and since ${{\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)}^{-\dfrac{1}{2}}}$ can be written as $\dfrac{1}{\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}}.$
$\Rightarrow \dfrac{\partial w}{\partial z}=\dfrac{z}{\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}}$
Hence, the derivative of $w=\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}$ is represented by, $\dfrac{\partial w}{\partial x}=\dfrac{x}{\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}}$ , $\dfrac{\partial w}{\partial y}=\dfrac{y}{\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}}$ , and $\dfrac{\partial w}{\partial z}=\dfrac{z}{\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}}.$

Note:
We need to know the basic formulae for partial differentiation. We need to be careful while partially differentiating with respect to a variable because we need to treat all the other variables as constants else it would lead to a wrong result.