Implicit Differentiation of 3 variables

Question

Find $\frac{dy}{dz}$ when

$$(-5x+z)^4-2x^3y^6+3yz^6+6y^4z=10.$$

I got an answer of

$$\frac{-24y^3(z-3)z^6+2x^36y^5}{4(-5x+z)^3+8yz^5+6y^4}.$$

Is this correct?

Answer 1

You are considering the equation: \begin{equation*} (-5x+z)^{4}-2x^{3}y^{6}+3yz^{6}+6y^{4}z=10\end{equation*} and you wish to calculate $\frac{dy}{dz}$. It follows that \begin{align*} 0&=\frac{d}{dz}{10}=\frac{d}{dz}\left[(-5x+z)^{4}-2x^{3}y^{6}+3yz^{6}+6y^{4}z\right] \\ &=4(z-5x)^{3}\left(1-5\frac{dx}{dz}\right)-2\left[6x^{3}y^{5}\frac{dy}{dz}+3x^{2}\frac{dx}{dz}y^{6}\right]+3\left[6yz^{5}+z^{6}\frac{dy}{dz}\right]+6\left[y^{4}+4zy^{3}\frac{dy}{dz}\right]\end{align*} Can you take it from here to verify whether or not your answer is correct?

Answer 2

Calculate the differential of the entire equation $$\eqalign{ 0 &= 6\,d(y^4z) - 2\,d(x^3y^6) +3\,d(yz^6) + d(z-5x)^4 \\ \\ 0 &= 6(y^4dz+4zy^3dy) - 2(3x^2y^6dx+6x^3y^5dy) \\ &\quad+\; 3(z^6dy+6yz^5dz) + 4(z-5x)^3(dz-5\,dx) \\ }$$ Assuming $x$ is held constant (i.e. $\,dx=0$) allows us to drop some terms $$\eqalign{ 0 &= (6y^4dz+24zy^3dy) - (12x^3y^5dy)+ (3z^6dy+18yz^5dz) + 4(z-5x)^3dz \\ 0 &= \Big(24zy^3+3z^6-12x^3y^5\Big)dy + \Big(6y^4+18yz^5+4(z-5x)^3\Big)dz \\ }$$ This can be rearranged into the desired partial derivative $$\eqalign{ \frac{\partial y}{\partial z} &= \frac{6y^4+18yz^5+4(z-5x)^3}{12x^3y^5-24zy^3-3z^6} \\ }$$