0 votes 0 votes Let $w=\log(u^2 +v^2)$ where $u=e^{(x^2+y)}$ and $v=e^{(x+y^2)}$. Then $\frac{\partial w }{\partial x} \mid _{x=0, y=0}$ is $0$ $1$ $2$ $4$ Others isi2015-mma partial-derivatives non-gate + – Arjun asked Sep 23, 2019 • recategorized Nov 17, 2019 by Lakshman Bhaiya Arjun 526 views answer comment Share Follow See 1 comment See all 1 1 comment reply NastyBall commented Jun 20, 2021 reply Follow Share Option B is the correct answer. 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes We can find the value of $\frac{\partial w}{\partial x}=\frac{\partial w}{\partial v}\times \frac{\partial v}{\partial x}$ $\therefore$ $\frac{\partial w}{\partial v}=$ $\frac{1}{u^{2}+v^{2}}\times \left ( 2v \right )$ ----$\left ( 1 \right )$ $\frac{\partial v}{\partial x}=e^{x+y^{2}}\times \left ( 1 \right )$ ---$\left ( 1 \right )$ Multipying and simplifying $\left ( 1 \right )$ and $\left ( 2 \right )$, we get $\frac{\partial w}{\partial x}=\frac{2}{u^{2}+v^{2}}\times \left ( v \right )\times e^{x+y^{2}}$ $\Rightarrow$ $\frac{\partial w}{\partial x}_{|x=0,y=0}$ $=$ $\frac{2\times 1}{1\times 1}\times 1=2$ $\therefore$ Option C is the answer. haralk10 answered Mar 14, 2020 haralk10 comment Share Follow See all 2 Comments See all 2 2 Comments reply rahulshah commented Aug 18, 2020 reply Follow Share @haralk10 You forgot to take the derivatives wrt u. Also, the answer is 1, because in the last step 1x1 is not equal to 1+1. Multiplication is usually not equal to addition unless both the operands are zero 0 votes 0 votes ksdiwe commented Sep 5, 2020 reply Follow Share answer is 1 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes Use chain rule, $\frac{\partial w}{\partial x}=\frac{\partial w}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial w}{\partial v}\frac{\partial v}{\partial x}$ Find $u,v$ at $x=0,y=0$. You will get $1$ as your answer. KAUNIL answered Mar 9, 2023 KAUNIL comment Share Follow See all 0 reply Please log in or register to add a comment.