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Let $z(x,y)$ be implicitly defined by the equation $3x+4y+4z+3cos(2z)+1=0$

I want to find $\frac{\partial^2 z}{\partial x \partial y}$.

My attempt: $\frac{\partial z}{\partial x} = 3+4\frac{\partial y}{\partial x} + 4\frac{\partial z}{\partial x}-6sin(2z)\frac{\partial z}{\partial x} = 0$

I'm not sure how to go about differentiating this expression once again, but now with respect to $y$. Is my first expression correct, atleast? And how do I proceed?

Thank you

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  • $\begingroup$ is here $$y=y(x)$$ also given? $\endgroup$ – Dr. Sonnhard Graubner Apr 12 '18 at 14:10
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Assuming $x$ and $y$ are independent, you can first differentiate w.r.t $x$

$$ 3 + 4\frac{\partial z}{\partial x} - 6\sin (2z)\frac{\partial z}{\partial x} = 0 $$

Then differentiate w.r.t $y$ (and using the product rule)

$$ 4\frac{\partial^2 z}{\partial x\partial y} -12\cos(2z)\frac{\partial z}{\partial y}\frac{\partial z}{\partial x} - 6\sin(2z)\frac{\partial^2 z}{\partial x\partial y} = 0 $$

Solving for the second partial yields $$ \frac{\partial^2 z}{\partial x\partial y} = \frac{12\cos(2z) \frac{\partial z}{\partial x}\frac{\partial z}{\partial y}}{4-6\sin(2z)} $$

If you want to express the first partials in terms of $x,y,z$, you can do it by differentiating the original equation w.r.t $y$ first $$ 4 + 4\frac{\partial z}{\partial y} - 6\sin(2z) \frac{\partial z}{\partial y} = 0 $$

Then $$ \frac{\partial z}{\partial x} = \frac{3}{6\sin (2z) - 4} $$ $$ \frac{\partial z}{\partial y} = \frac{4}{6\sin(2z)-4} $$

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