The partial derivatives of $F(x,y)=x^3y^4-\ln (x^2y)+e^{3x-6y}$

Question

This is what I have so far, I am not really sure what I am doing is correct. Can anyone help with this?

Given $F(x,y)=x^3y^4-\ln (x^2y)+e^{3x-6y}$

Find $f_x,f_y,f_{xx},f_{xy},f_{yy}$

$$P_x=3x^2y^4-\frac{2xy}{xy}+e^{x-6y}$$ $$P_{xx}=6xy^4-2+e^{-6y}$$ $$P_y=x^34y^3-\frac{x^2}{x^2}+e^{3x-6}$$ $$P_x=x^312y^2-1+e^{3x}$$ $$P_{xy}=\frac{6xy^4-2+e^{-6y}}{x^3-12y^2-1+e^{3x}}$$

Answer 1

There is nothing to be worried about even when partially differentiating with respect to an independent variable.

NOTE: When you partially differentiate with respect to $x$, treat $y$ as constant.

I'll solve the first one for you as an example

$(x^3y^4)_x=3x^2y^4$

$(\ln(x^2y))_x={1\over x^2y}\cdot(x^2y)_x={2xy\over x^2y}$ (You missed the square in the denominator)

$(e^{3x-6y})_x=e^{3x-6y}\cdot (3x-6y)_x=3e^{3x-6y}$

Now club the three.

Answer 2

You did good. There are a few mistakes though

$$P_x = 3x^2y^4 - \frac{2}{x} + 3e^{3x-6y}$$

$$P_y = 4x^3y^3 - \frac{1}{y} - 6e^{3x-6y}$$

$$P_{xx} = 6xy^4 + \frac{2}{x^2} + 9e^{3x-6y}$$

Keep going for $P_{xy}$ and $P_{yy}$.

If you're differetiating w.r.t. $x$, remember to treat $y$ as a constant. For instance, differetiating w.r.t. $x$ and also using the chain rule we have

$$\frac{\partial \ln (x^2y)}{\partial x} = \frac{2xy}{x^2y} = \frac{2}{x}$$