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

  • $\begingroup$ is here $$y=y(x)$$ also given? $\endgroup$ – Dr. Sonnhard Graubner Apr 12 '18 at 14:10

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.