Finding $\frac{\partial z}{\partial x}$ when sin(5x-4y+z)=0

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sin(5x-4y+z)=0

how do I find \frac{\partial z}{\partial x}?

if the problem is sin(5x-4y+z)=f(x,y,z), I can find \frac{\partial f}{\partial x} but I don't know what to do when it is just equal to zero.
 
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Solve for z, take the partial derivative.
 
Simpler: take the partial derivative with respect to x, assuming that y is independent of x, z a function of x, then solve for zx. Use the chain rule. Remember "implicit differentiation" from Calculus I?
(sin(5x- 4y+ z))x= cos(5x- 4y+ z)(5- zx)= 0. Solve for zx.
 
HallsofIvy: I think that should be a (5+zx) in your second expression.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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