consider the following function: f(x, y)=(4x + 3y)^(3/5) step 1 of 3: find f_xx. answer f_xx =

Explanation:

Step1: Use chain - rule for first - order derivative

Let $u = 4x+3y$, then $f(x,y)=u^{\frac{2}{3}}$. The partial derivative of $f$ with respect to $x$ is $f_x=\frac{\partial f}{\partial u}\cdot\frac{\partial u}{\partial x}$. $\frac{\partial f}{\partial u}=\frac{2}{3}u^{-\frac{1}{3}}$ and $\frac{\partial u}{\partial x}=4$, so $f_x=\frac{2}{3}(4x + 3y)^{-\frac{1}{3}}\cdot4=\frac{8}{3}(4x + 3y)^{-\frac{1}{3}}$.

Step2: Use chain - rule for second - order derivative

To find $f_{xx}$, we differentiate $f_x$ with respect to $x$ again. Let $v=(4x + 3y)^{-\frac{1}{3}}$, and $f_x=\frac{8}{3}v$. The derivative of $v$ with respect to $x$ is $\frac{\partial v}{\partial x}=-\frac{1}{3}(4x + 3y)^{-\frac{4}{3}}\cdot4$. Then $f_{xx}=\frac{8}{3}\cdot\left(-\frac{4}{3}(4x + 3y)^{-\frac{4}{3}}\right)=-\frac{32}{9}(4x + 3y)^{-\frac{4}{3}}$.

Answer:

$-\frac{32}{9}(4x + 3y)^{-\frac{4}{3}}$