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# Can someone please explain how you calculate $\dfrac{{df}}{{dx}}$ ? This is the function $f(x) = \sqrt {1 - ({x^2} + {y^2})}$ ?

Last updated date: 11th Jun 2024
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Hint: Differentiation is known as the process of dividing a whole quantity into very small ones. In this question, a function is given to us that involve the square root of x raised to some power so the given function is in terms of x, we have to differentiate $f(x) = \sqrt {1 - ({x^2} + {y^2})}$ with respect to x. So, the independent variable is x and the dependent variable is y.

We will first differentiate the whole quantity $f(x) = \sqrt {1 - ({x^2} + {y^2})}$ and then differentiate the quantity in the square root as it is also a function of x $[1 - ({x^2} + {y^2})]$ . The result of multiplying these two differentiated functions will give the value of $\dfrac{{df}}{{dx}}$ or $f'(x)$ .On solving the given question using the above information, we will get the correct answer.

Complete step-by-step solution:
We are given $f(x) = \sqrt {1 - ({x^2} + {y^2})}$
We know that $\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}$
So differentiating both sides of the above equation with respect to x, we get –
$\Rightarrow \dfrac{{df}}{{dx}} = \dfrac{1}{2}{[1 - ({x^2} + {y^2})]^{ - \dfrac{1}{2}}}\dfrac{{d[1 - ({x^2} + {y^2})]}}{{dx}} \\ \Rightarrow \dfrac{{df}}{{dx}} = \dfrac{1}{{2\sqrt {1 - ({x^2} + {y^2})} }}( - 2x) \\ \Rightarrow \dfrac{{df}}{{dx}} = \dfrac{{ - x}}{{\sqrt {1 - ({x^2} + {y^2})} }} \\$
Hence, the $\dfrac{{df}}{{dx}}$ of the function $f(x) = \sqrt {1 - ({x^2} + {y^2})}$ is $\dfrac{{ - x}}{{\sqrt {1 - ({x^2} + {y^2})} }}$.

Note: We use differentiation when we have to find the instantaneous rate of change of a quantity, it is represented as $\dfrac{{dy}}{{dx}}$ , in the expression $\dfrac{{dy}}{{dx}}$ , a very small change in quantity is represented by $dy$ and the small change in the quantity with respect to which the given quantity is changing is represented by $dx$ . The given function contains more than one variable, that is, it is a multivariable equation. So, when we differentiate the function with respect to one variable, we treat the other variable as constant.