Question 11.2.5: If f (x, y) = x³e^y², find the first- and second-order parti......
If f (x, y) = x^{3}e^{y^{2}}, find the first- and second-order partial derivatives at the point (x, y) = (1, 0).
To find f^{\prime}_{1}(x, y), we differentiate x^{3}e^{y^{2}} w.r.t. x while treating y as a constant. When y is a constant, so is e^{y^{2}} . Hence, f^{\prime}_{1}(x, y) = 3x^{2}e^{y^{2}} and so
f_{1}^{\prime}(1,0)=3\cdot1^{2}e^{0^{2}}=3To find f^{\prime}_{2}(x, y), we differentiate f (x, y) w.r.t. y while treating x as a constant:
f_{2}^{\prime}(x,y)=x^{3}2y e^{y^{2}}=2x^{3}y e^{y^{2}}and so f^{\prime}_{2}(1, 0) = 0.
To find the second-order partial f^{\prime\prime}_{11}(x, y), we must differentiate f^{\prime}_{1}(x, y) w.r.t. x once more, while treating y as a constant. Hence, f^{\prime\prime}_{11}(x, y) = 6xe^{y^{2}} and so
To find f^{\prime\prime}_{22}(x, y), we must differentiate f^{\prime}_{2}(x, y) = 2x^{3}ye^{y^{2}} w.r.t. y once more, while treating x as a constant. Because ye^{y^{2}} is a product of two functions, each involving y, we use the product rule to obtain
f_{22}^{\prime\prime}(x,y)=(2x^{3})(1\cdot e^{y^{2}}+y2y e^{y^{2}})=2x^{3}e^{y^{2}}+4x^{3}y^{2}e^{y^{2}}Evaluating this at (1, 0) gives f^{\prime\prime}_{22}(1, 0) = 2. Moreover,
f_{12}^{\prime\prime}(x,y)=\frac{\partial}{\partial y}\left[f_{1}^{\prime}(x,y)\right]=\frac{\partial}{\partial y}(3x^{2}e^{y^{2}})=3x^{2}2y e^{y^{2}}=6x^{2}y e^{y^{2}}and
f_{21}^{\prime\prime}(x,y)=\frac{\partial}{\partial x}\left[f_{2}^{\prime}(x,y)\right]=\frac{\partial}{\partial x}(2x^{3}y e^{y^{2}})=6x^{2}y e^{y^{2}}Hence, f^{\prime\prime}_{12}(1, 0) = f^{\prime\prime}_{21}(1, 0) = 0.