Question Find the second‑order Taylor formula for 𝑓(𝑥,𝑦)=5sin(𝑥𝑦)+3cos(𝑥𝑦) at (0,0). (Use symbolic notation and fractions where needed. Give your answer in the form of 𝑓(ℎ1,ℎ2)=𝑓(𝑙,𝑚) where 𝑙=ℎ1 and 𝑚=ℎ2.)

Solution \Rightarrow T_{0} solve this question, we use following definition For a function of two variables f(x, y) whose first and second Partial exist at the Point (a, b), then and degree Taylor Polynomial of f(r, y) neer the Point (a, b) is\begin{aligned}f(x, y)= & \theta(x, y)=f(a, b)+f_{x}(0, b)(x-a)+f_{y}(a, b)(y-b) \\& +\frac{f_{x x}(a, b)}{2}(x-a)^{2}+f_{x y}(a, b)(x-a)(y-b)+ \\& \frac{f_{y y}(a, b)}{2}(y-b)^{2}\end{aligned}Here, given f(x, y)=5 \sin (x y)+3 \cos (x y) at 10,0\begin{array}{l}\therefore f(a, b)=3=f(0,0) \quad\left\{\begin{array}{ll}\therefore & \cos 0=1\end{array}\right\} \\=\frac{\partial}{\partial x}[5 \sin (x y)+3 \cos (x y)] \\=5 \cos (x y) \cdot \frac{\partial}{\partial x}(x y)+3(-\sin (x y)) \frac{\partial}{\partial x}(x y) \\=5 \cos (x y) \cdot y \cdot 1-3 \sin (x y) \cdot y \cdot 1 \\f_{x}(x, y)=[5 \cos (x y)-3 \sin (x y)] y \\f_{x}(0,3)=f_{x}(0,0)=[5 \cos 0-3 \sin D] \cdot 0 \\\therefore f_{x}(0,0)=0 \\\end{array}\begin{aligned}f_{y}(x, y) & =\text { Partial derivative of } f(x, y) \text { w.x.t.y } \\& =\frac{\partial}{\partial y}[5 \sin (x y)+3 \cos (x y)] \\& =5 \cos (x y) \frac{\partial}{\partial y}(x y)-3 \sin (x y) \frac{d}{\partial y}(x y)\end{aligned}\begin{array}{l}=5 \cos (x y) \cdot x \cdot 1-3 \sin (x y) \cdot x \cdot 1 \\f_{y}(x, y)=\left[5 \cos (x y)-3 \sin ^{\prime \prime \prime}(x y)\right] x \\\text { Now } \begin{aligned}\text { fy }(a, b)=f_{y}(0,0)= & {[5 \cos D-3 \sin D] \cdot 0 } \\& =0\end{aligned} \\\text { SD } f_{y}(a, b)=D=f_{y}(0,0)-(3) \\=\frac{\partial}{\partial x}\left(f_{x}(x, y)\right) \\=\frac{\partial}{\partial x}[5 \cos (x y) \cdot y-3 \sin (x y) y][\text { fiom (I) }] \\=5 y \frac{\partial}{\partial x} \cos (x y)-3 y \frac{\partial}{\partial x}(\sin (x y)) \\=-5 y \sin (x y) \cdot \frac{\partial}{\partial x}(x y)-3 y \cos (x y) \frac{\partial}{\partial x}(x y) \\=-5 y \sin (x y) \cdot y-3 y \cos (x y) \cdot y \\f_{x x}(x, y)=-5 y^{2} \sin (x y)-3 y^{2} \cos (x y) \\\begin{aligned}f_{x x}(a, b)=f_{x x}(0,0) & =-5 \cdot 0 \cdot \sin 0-3 \cdot 0 \cdot \cos 0 \\& =0\end{aligned} \\\end{array}Now f_{x_{x}}(x, y)= Second derivative of f(x, y)= w.1.t. x = again differentiate f_{x}(x, y) w.x.t. xSo f_{x_{x}}(a, b)=0=h_{\left.x_{x}, 0,0\right)}(4)\begin{aligned}\text { Now for }(x, y) & =\text { Second derivative of } f(x, y) \text { w.l.t } y \\& =\text { again differentiate } f_{y}(x, y) \text { w.r.t. } y \\& =\frac{\partial}{\partial y}\left[f_{y}(x, y)\right] \\& =\frac{\partial}{\partial y}[5 \cos (x y) x-3 \sin (x, x] \quad[\text { using (ii) }] \\& =5 x \frac{\partial}{\partial y} \operatorname{les}(x y)-3 x \frac{\partial}{\partial y} \sin (x y) \\& =-5 x \sin (x y) \frac{\partial}{\partial y}(x y)-3 x \cos (x y) \frac{\partial}{\partial y}(x y)\end{aligned}\begin{array}{l}=-5 x \sin (x y) \cdot x-3 x \cos (x y) \cdot x \\ \therefore f_{y y}\left(x_{y}\right)=-5 x^{2} \sin (x y)-\cdots x^{2} \cos (x y) \\ \begin{aligned} f_{y y}(a, b)=f_{y y}(0,0) & =-5 \cdot 0 \cdot \sin 0-3 \cdot 0 \cdot \cos 0 \\ & =0\end{aligned} \\ \therefore b_{y y}(a, b)=0 \\ =f_{y y}(0,0) \\ \begin{aligned} f_{x y}(x, y) & =\text { derivative of } f_{x}(x, y) \text { w.x.t. } y=\frac{\partial^{2} f}{\partial y \partial x} \\ & =\frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x}\right)\end{aligned} \\ =\frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x}\right) \\ =\frac{\partial}{\partial y}\left(f_{x}\right) \\ =\frac{\partial}{\partial y}[5 \cos (x y) y-3 \sin (x y) y] \\ =\left\{5 \cos (x y) \cdot \frac{\partial}{\partial y}(y)+y \frac{\partial}{\partial y} \cos (x y)\right\} \\ -\left\{3 \sin (x y) \frac{\partial}{\partial y}(y)+y \cdot \frac{\partial}{\partial y} \sin (x y)\right\} \text {. } \\ {\left[\begin{array}{l}\text { using product Rule of desinatives } \\ \frac{d}{d x}(U v)=U \cdot \frac{d}{d x} v+v \cdot \frac{d}{d x} v\end{array}\right]} \\ =5 \cos (x y) \cdot 1+y[-\sin (x y)] \cdot \frac{\partial}{\partial y}(x y) \\ -3 \sin (x y) \cdot 1-y \cdot \cos (x y) \cdot \frac{\partial}{\partial y}(x y) \\ =5 \operatorname{Cos}(x y)-y \operatorname{Sin}(x y) \cdot x-3 \operatorname{Sin}(x y)-y \operatorname{Cos}(x y) \cdot x \\ f_{x y}(x, y)-5 \cos (x y)-x y \operatorname{Sin}(x y)-3 \sin (x y)-x y \cos (x y) \\ \text { fxy }(a, b)=f_{x y}(0,0)=56 \cos 0-0-3 \sin 0-0 \\ =5 \cdot 1-0-3 \cdot 0-0 \\ =5-3=2 \quad[\because \sin \theta=0 \\ \therefore \text { fxy }(a, b)=2-(6) \\ \text { ceso }=1 \\ -f_{x y}(0,0) \\\end{array}So 2nd dogex taylor polynomial of f(x, y) neer (0,0) us given by\begin{aligned}f(x, y)= & \theta(x, y)=f(0,0)+f_{x}(0,0)(x-0)+f_{y}(0,0)(y-0) \\& +\frac{f_{x x}(0,0)}{2}(x-0)^{2}+f_{x y}(0,0)(x-0)(y-0)\end{aligned}+\frac{f_{y y}(0,0)}{2}(y-0)^{2} \quad[\text { using }(A)]=3+0 \cdot x+0 \cdot y+\frac{0}{2} x^{2}+2 \cdot x \cdot y+\frac{0}{2} y^{2}[ using (1),(2),(3),(4,(5),(6)]=3+0+0+0+2 x y+0\therefore Q(x, y)=3+2 x y \text { Ans. } ...