ISI2015-MMA-70

Question

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

• A. $0$
• B. $1$
• C. $2$
• D. $4$

Comments

Option B is the correct answer.

Answer 1

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.

Comments

@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

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answer is 1

Answer 2

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.