Derivative of $x^{y^z} + y^{x^z}$ with respect to $z$

Question

Please what is the partial derivative of $x^{y^z} + y^{x^z}$ with respect to $z$ I guess this is not ${(x^y)}^z$ but rather $x^{(y^z)}$ so I tried using the logarithm function and found $\ln x\ln y(x^{y^z} y^z+y^{x^z} x^z)$ But I don't know if I got it correctly

Any help would be appreciated thanks

Answer 1

Write $$x^{y^z}=e^{\log(x^{y^z})}=e^{y^z \log x}=e^{\log x e^{z ]log y}}\\ \frac {\partial (e^{\log x e^{z ]log y}})}{\partial z}= e^{\log x e^{z ]log y}}\log xe^{z \log y}\log y$$

Answer 2

Note for $u(z) =a^{f(z)}$, or $\ln u(z) =f(z)\ln a$,

$$u’(z) =u(z) f’(z) \ln a =a^{f(z)}f’(x) \ln a$$

Then, apply,

$$(x^{y^z} + y^{x^z} )’$$ $$=x^{y^z} (y^z)’\ln x + y^{x^z} (x^z)’ \ln y$$ $$=x^{y^z}y^z\ln x\ln y + y^{x^z}x^z\ln y \ln x $$ $$=(x^{y^z}y^z+ y^{x^z}x^z)\ln x \ln y $$

So, you have it right.