### CSIR JUNE 2011 PART C QUESTION 69 SOLUTION (The derivative of $f(x,y) = (3x - 2y + x^2, 4x +5y + y^2)$)

Consider the function $f: \Bbb R^2 \to \Bbb R^2$ defined by $f(x,y) = (3x - 2y + x^2, 4x +5y + y^2)$. Then
1) f is discontinuous at $(0,0)$,
2) $f$ is continuous at $(0,0)$ and all the directional derivatitives exist at $(0,0)$,
3) $f$ is differentaible at $(0,0)$ but the derivative $Df(0,0)$ is not invertible,
4) $f$ is differentaible at $(0,0)$ and the derivative $Df(0,0)$ is invertible.
Solution: Let $f_1(x,y) = 3x-2y+x^2$ and $f_2 = 4x+5y+y^2$ be the coordinate functions of $f$.
The partial derivatives of $f$ are $\frac{\partial f_1}{\partial x} = 3 + 2x$, $\frac{\partial f_1}{\partial y}= -2$, $\frac{\partial f_2}{\partial x} = 4$ and $\frac{\partial f_2}{\partial y} = 5 + 2y$.
option 1:(True) The coordinate functions $f_1$ and $f_2$ of $f$ are polynomials and hence $f$ is continuous. This can also be seen from the continuity of partial derivativatives of the coordinate functions of $f$.
option 3: (False) The derivative of a function $f: \Bbb R^n$ to $\Bbb R^m$ at a point $x \in \Bbb R^n$, if it exists, is the unique linear transformation $D(f(x)) \in \mathcal{L}(\Bbb R^n,\Bbb R^m)$ such that $$\text{lim}_{h \to 0} \frac{||f(x+h) - f(h) - D(f(x))h||}{||h||} \to 0.$$ where
$$Df(x) = \begin{bmatrix} \frac{\partial f_1}{\partial x}&\frac{\partial f_1}{\partial y} \\ \frac{\partial f_2}{\partial x}&\frac{\partial f_2}{\partial y}\end{bmatrix} =\begin{bmatrix}3+2x&-2\\4&5+2y\end{bmatrix}$$ and $Df((0,0)) = \begin{bmatrix}3&-2\\4&5\end{bmatrix} .$

Let $x = \begin{bmatrix}0 \\ 0 \end{bmatrix}$ and $h = \begin{bmatrix}h_1 \\ h_1 \end{bmatrix}$ then $f(x+h) = f(h)$. The above limit becomes $$\text{lim}_{h \to 0} \frac{||- D(f((0,0)))\cdot h||}{||h||} = \text{lim }_{h \to 0}\frac{||\begin{bmatrix}3h_1-2h_2 \\ 4h_1+5h_2 \end{bmatrix}||}{||\begin{bmatrix}h_1 \\ h_2 \end{bmatrix}||}$$
By calculating the norm explicitly, we notice that, the above limit exists. Hence $Df((0,0)) = \begin{bmatrix}3&-2\\4&5\end{bmatrix}$ is the required derivative of $f$ at $(0,0)$ which is clearly invertible and option (4) is true and in turn option (3) is false . Since $f$ is differential at $(0,0)$, we have option (2) is also correct.

### NBHM 2020 PART A Question 4 Solution $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$
Evaluate : $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$ Solution : \int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx = \int_{-\infty}^{\inft...