Finding the partial derivatives of $V (x, y) = U (x, y)e^{−ax−by}$

Question

I think I did something wrong, so I was hoping someone might be able to show me the solution

Two functions $V (x, y)$ and $U (x, y)$ are connected by the equation

$$V (x, y) = U (x, y)e^{−ax−by}$$

where $a$ and $b$ are constants.

a) Find $$\frac{\partial V}{\partial x},\,\frac{\partial V}{\partial y},\,\frac{\partial^2 V}{\partial x^2},\,\frac{\partial^2 V}{\partial y^2}$$

in terms of $U$ and its partial derivatives.

b) Suppose that $V$ satisfies

$$\frac{\partial V}{\partial y}=\frac{\partial^2 V}{\partial x^2}+2\frac{\partial V}{\partial x}-3V$$

c) If $a = 1$ and $b = 4$, show that the function $U$ then satisfies the equation $$\frac{\partial U}{\partial y}=\frac{\partial^2 U}{\partial x^2}$$

Progress

I have $$V_x = U_x -a e^{-ax-by}$$ $$V_y = U_y -b e^{-ax-by}$$ $$V_{xx}= U_{xx}-a e^{-ax-by}$$ which of course doesn't make sense.

Answer 1

You remembered to use the chain rule, but apparently forgot the product rule. Use it once: $$ V_x=e^{-ax-by}U_x-a e^{-ax-by}U $$ then use it again: $$ V_{xx}=e^{-ax-by}U_{xx}- a e^{-ax-by}U_x - a e^{-ax-by}U_x + a^2 e^{-ax-by}U $$ You may notice the same term appearing in the middle twice; it saves a bit of time to remember the product formula for second derivative: $$ (uv)'' = u''v+2u'v'+uv'' $$ which is -- not coincidentally -- similar to the binomial formula.

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Using subscripts for derivatives, e.g., $U_x$ instead of $\frac{\partial U}{\partial x}$, speeds up the computation as well.

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I trust that after your derivatives of $V$ are in order, plugging them into (b) to get (c) will not be too difficult.

Answer 2

Let $\alpha=c_1$, $\beta=c_2$ and $U=f$(completely for shortening the latex solution, no real need to do that)$\frac{d}{dx}(f(x, y)e^{-(c_1 x)-c_2 y)}$ $\cdot \frac{d}{dx}(uv) = v \frac{du}{ dx}+u\frac{dv}{dx}$, where $u = e^{c_1 (-x)-c_2 y}$ and $v = f(x, y)$

$=f(x, y) (\frac{d}{dx}(e^{-(x c_1)-y c_2}))+e^{-x c_1-y c_2} (\frac{d}{dx}(f(x, y)))$ $\cdot$ (Chain rule) $e^{-(x c_1)-y c_2}(\frac{d}{dx}(f(x, y)))+e^{-(x c_1)-y c_2} \frac{d}{dx}(-(x c_1)-y c_2) f(x, y)=e^{-(x c_1)-y c_2} (\frac{d}{dx}(f(x, y)))+\frac{d}{dx}(-(y c_2))-\frac{d}{dx}(x) c_1 e^{-(x c_1)-y c_2} f(x, y)$ The derivative of $c_2 (-y)$ is zero(we're differentiating with respect to x so we treat $y$ as a constant) $= e^{-(x c_1)-y c_2} (\frac{d}{dx}(f(x, y)))+e^{-(x c_1)-y c_2} f(x, y) (0-(\frac{d}{dx}(x)) c_1) = e^{-(x c_1)-y c_2} (\frac{d}{dx}(f(x, y)))-e^{-(x c_1)-y c_2} f(x, y) (\frac{d}{dx}(x)) c_1=f_x' e^{-(x c_1)-y c_2}-e^{-(x c_1)-y c_2} f(x, y) (\frac{d}{dx}(x)) c_1 = -e^{-(x c_1)-y c_2} f(x, y) c_1+e^{-(x c_1)-y c_2} f_x$. The derivative of $f(x,y)$ wrt x is $f_x$.

I hope you got the intuition for all the rest.