Derivative of a Function with Respect to x using Quotient Rule and Chain Rule
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.
