Q.If $V(x,y)=e^x(x\cos y-y\sin y)$, then prove that $\dfrac{\partial^2 V}{\partial x^2}+\dfrac{\partial^2 V}{\partial y^2}=0$.

Harmonic is, by definition, the property of satisfying Laplace's equation at every point of the domain.

1. A twice continuously differentiable function $u:A\to R$ is called harmonic on the open set $A$ if it satisfies Laplace's partial differential equation $\dfrac{\partial^2u}{\partial x^2}+\dfrac{\partial^2u}{\partial y^2}=0$ at every point of $A$.
2. This is the standard definition used throughout the theory of harmonic functions in two variables — no alternative (nonzero, subtraction-based) form defines harmonicity. …

Applying the chain rule to the exponential with $y$ held fixed brings down a factor of $2x$, reproducing $u$ itself.

1. $u(x,y)=e^{x^2+y^2}$. Differentiate partially with respect to $x$, treating $y$ as constant.
2. By the chain rule: $\dfrac{\partial u}{\partial x}=e^{x^2+y^2}\cdot\dfrac{\partial}{\partial x}\left(x^2+y^2\right)$. …

Apply the chain rule to $u=e^{x^2}$: $\dfrac{\partial u}{\partial x}=2x\,e^{x^2}$.

Here $u=e^{x^2}$ is a function of $x$ (treated as a function of $x$ for the partial derivative). Using the chain rule, $\dfrac{d}{dx}\,e^{f(x)}=e^{f(x)}\cdot f'(x)$ with $f(x)=x^2$:

$\frac{\partial u}{\partial x}=e^{x^2}\cdot\frac{d}{dx}(x^2)=e^{x^2}\cdot 2x=2x\,e^{x^2}.$

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$\dfrac{\partial q}{\partial p_1}=8$.

To find the partial derivative with respect to $p_1$, treat $p_2$ (and any constant) as fixed and differentiate term by term:

$q=1000+8p_1-p_2.$

• $\dfrac{\partial}{\partial p_1}(1000)=0$ (constant),
• $\dfrac{\partial}{\partial p_1}(8p_1)=8$, …

Applying the chain rule to $u = e^{x^2}$ gives $\dfrac{\partial u}{\partial x} = 2x\,e^{x^2}$, so the answer is option (b).

Here $u = e^{x^2}$. Differentiate partially with respect to $x$ using the chain rule $\dfrac{d}{dx}e^{g(x)} = e^{g(x)}\,g'(x)$ with $g(x) = x^2$:

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Differentiating $u=e^{x^2+y^2}$ partially with respect to $x$ using the chain rule gives $\dfrac{\partial u}{\partial x}=2xu$.

1. We are given $u(x,y)=e^{x^2+y^2}$.
2. To find $\dfrac{\partial u}{\partial x}$, differentiate with respect to $x$ while treating $y$ as a constant.
3. Let $w=x^2+y^2$, so $u=e^{w}$. By the chain rule, $\dfrac{\partial u}{\partial x}=e^{w}\cdot\dfrac{\partial w}{\partial x}$.
4. Since $y$ is held constant, $\dfrac{\partial w}{\partial x}=2x$. …

The partial derivative of $u=e^{x^2}$ with respect to $x$ is $2x\,e^{x^2}$.

Given $u = e^{x^2}$. Differentiate partially with respect to $x$ (chain rule: derivative of $e^{f}$ is $e^{f}\cdot f'$): …

When $u(x,y)$ is continuous (with continuous partials), the mixed second-order partial derivatives are equal, so $\dfrac{\partial^2 u}{\partial y\,\partial x}=\dfrac{\partial^2 u}{\partial x\,\partial y}$.

The symbol $\dfrac{\partial^2 u}{\partial y\,\partial x}$ means: first differentiate $u$ partially with respect to $x$, then with respect to $y$. The symbol $\dfrac{\partial^2 u}{\partial x\,\partial y}$ reverses the order.

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