Find $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ for $f(x, y)=6 x^{3}+8 y^{2}-4 x+7 y$ and evaluate each at $(3,4)$ \[ \begin{array}{l} \frac{\partial f}{\partial x}= \\ \left.\frac{\partial f}{\partial x}\right|_{(3,4)}= \\ \frac{\partial f}{\partial y}= \\ \left.\frac{\partial f}{\partial y}\right|_{(3,4)}= \end{array} \]
Solution
Step 1 ：Differentiate the function \(f(x, y)=6 x^{3}+8 y^{2}-4 x+7 y\) with respect to \(x\) treating \(y\) as a constant to get \(\frac{\partial f}{\partial x} = 18x^{2} - 4\).
Step 2 ：Differentiate the function \(f(x, y)=6 x^{3}+8 y^{2}-4 x+7 y\) with respect to \(y\) treating \(x\) as a constant to get \(\frac{\partial f}{\partial y} = 16y + 7\).
Step 3 ：Evaluate \(\frac{\partial f}{\partial x}\) at the point \((3,4)\) to get \(\left.\frac{\partial f}{\partial x}\right|_{(3,4)} = 18(3)^{2} - 4 = 158\).
Step 4 ：Evaluate \(\frac{\partial f}{\partial y}\) at the point \((3,4)\) to get \(\left.\frac{\partial f}{\partial y}\right|_{(3,4)} = 16(4) + 7 = 71\).
Step 5 ：So, the final results are \(\boxed{\frac{\partial f}{\partial x} = 18x^{2} - 4, \left.\frac{\partial f}{\partial x}\right|_{(3,4)} = 158}\) and \(\boxed{\frac{\partial f}{\partial y} = 16y + 7, \left.\frac{\partial f}{\partial y}\right|_{(3,4)} = 71}\).
