Question - Derivative of a Function with Respect to x using Quotient Rule and Chain Rule
Solution:
The image shows a mathematical problem asking to find the derivative with respect to $$x$$ of the function $$f(x,y)$$, which is provided as:$$\frac{d}{dx} \left( \frac{2xy}{\sqrt{x^2 + y^2}} \right)$$To find this derivative, we will use the quotient rule along with the chain rule because the function has both $$x$$ and $$y$$ variables, where $$y$$ could be a function of $$x$$.The quotient rule is:$$\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$$where $$u = 2xy$$ and $$v = \sqrt{x^2 + y^2}$$.The derivative of $$u = 2xy$$ with respect to $$x$$ is $$u' = 2y + 2xy'$$ (using the product rule and assuming $$y$$ is a function of $$x$$).The derivative of $$v = \sqrt{x^2 + y^2}$$ with respect to $$x$$ is $$v' = \frac{1}{2\sqrt{x^2 + y^2}} \cdot (2x + 2yy')$$ (using the chain rule and assuming $$y$$ is a function of $$x$$). Simplifying gives $$v' = \frac{x + yy'}{\sqrt{x^2 + y^2}}$$.Now, apply the quotient rule:$$\frac{d}{dx} \left( \frac{2xy}{\sqrt{x^2 + y^2}} \right) = \frac{(2y + 2xy')\sqrt{x^2 + y^2} - 2xy\left(\frac{x + yy'}{\sqrt{x^2 + y^2}}\right)}{x^2 + y^2}$$Simplify the expression:$$= \frac{2y(x^2 + y^2) + 2xy'y\sqrt{x^2 + y^2} - 2x^2y - 2xy^2y'}{(x^2 + y^2)^{3/2}}$$Now, combine like terms and simplify further:$$= \frac{2y^3 + 2xy'y\sqrt{x^2 + y^2} - 2xy^2y'}{(x^2 + y^2)^{3/2}}$$If $$y$$ is not a function of $$x$$, $$y'$$ would simply be 0, and the expression would be further simplified. However, the derivative expressed as it stands takes into consideration that $$y$$ may be a function of $$x$$. If you have more specific information about $$y$$, such as whether it is a constant or a function of $$x$$, you can simplify the expression accordingly.
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