Say I want to find the derivative of the below function with respect to x, which equals zero.
$u(x,z) = \ln(\frac{\mathrm{x}^α + \mathrm{z}^{α}}{\mathrm{z}^α}),$$ $$z = h(x) = \mathrm{(a\mathrm{x}^4 + b)}^{(1/3)}$.
$u'(x,z) = 0$.
As x and z are both functions of x, I expect to be able to solve this using the chain rule for multiple variables, however I always end up with the incorrect answer unless I substitute z.
$u'(x,z) = \frac{du}{dx} dx + \frac{du}{dz} dz = 0$.
By using the quotient rule on the logarithm, I end up with the wrong answer according to the textbook unless I substitute z then differentiate. How do you find the total derivative with the quotient rule? Is this even possible?
According to the answer: $x = \mathrm{(3b/a)}^{1/4}$.
I tried:
$u'(x,z) = \Large \frac{\frac{α\cdot\ \mathrm{z}^α\mathrm{x}^{α-1} - (0)}{\mathrm{z}^{α^{2}}} + \frac{\mathrm{z}^α α\mathrm{{z}^{α-1} -(\mathrm{x}^α + \mathrm{z}^{α}) α\cdot\mathrm{z}^{α-1}}} {\mathrm{z}^{α^{2}}}\frac{du}{dz}}{\frac{\mathrm{x}^α + \mathrm{z}^{α}}{\mathrm{z}^α}}$
Let $u(x,z)=\ln\left(\frac{x^\alpha + z^\alpha}{z^\alpha}\right) = \ln(x^\alpha + z^\alpha) - \ln(z^\alpha), z=h(x)=(ax^4 + b)^{1/3}$ . Then $$\frac{du}{dx}=\frac{\partial u}{\partial x}+\frac{\partial u}{\partial z}\frac{\partial z}{\partial x}$$ $$= \frac{\alpha x^{\alpha - 1}}{x^\alpha + z^\alpha} + \left(\frac{\alpha z^{\alpha - 1}}{x^\alpha + z^\alpha} - \frac{\alpha}{z}\right)\left(\frac{4 a x^3}{3 \left(a x^4+b\right)^{2/3}}\right)$$ $$=\frac{\alpha x^{\alpha -1} }{3 \left(x^{\alpha }+z^{\alpha }\right)}\left(3-\frac{4 a x^4}{z \left(a x^4+b\right)^{2/3}}\right)$$ $$=\frac{\alpha x^{\alpha -1}}{3 \left(\left(a x^4+b\right)^{\alpha /3}+x^{\alpha }\right)}\left(3-\frac{4 a x^4}{a x^4+b}\right)$$ Now if this is set to $0$, we get that $$3-\frac{4 a x^4}{a x^4+b}=0 \implies ax^4=3b \implies x=\left(\frac{3b}{a}\right)^{1/4}$$ This is why I asked if you copied everything correctly, since my answer is coming out to be different form the one you provided.