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Question: Can someone please explain how you calculate \(\dfrac{{df}}{{dx}}\) ? This is the function \(f(x) = ...

Can someone please explain how you calculate dfdx\dfrac{{df}}{{dx}} ? This is the function f(x)=1−(x2+y2)f(x) = \sqrt {1 - ({x^2} + {y^2})} ?

Explanation

Solution

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)=1−(x2+y2)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)=1−(x2+y2)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−(x2+y2)][1 - ({x^2} + {y^2})] . The result of multiplying these two differentiated functions will give the value of dfdx\dfrac{{df}}{{dx}} or f′(x)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)=1−(x2+y2)f(x) = \sqrt {1 - ({x^2} + {y^2})}
We know that dxndx=nxn−1\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}
So differentiating both sides of the above equation with respect to x, we get –
⇒dfdx=12[1−(x2+y2)]−12d[1−(x2+y2)]dx ⇒dfdx=121−(x2+y2)(−2x) ⇒dfdx=−x1−(x2+y2)  \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 dfdx\dfrac{{df}}{{dx}} of the function f(x)=1−(x2+y2)f(x) = \sqrt {1 - ({x^2} + {y^2})} is −x1−(x2+y2)\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 dydx\dfrac{{dy}}{{dx}} , in the expression dydx\dfrac{{dy}}{{dx}} , a very small change in quantity is represented by dydy and the small change in the quantity with respect to which the given quantity is changing is represented by dxdx . 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.