Let $z = x ^ { y }$ . Then the differential $d ^ { z }$ is

A) $\left( y x ^ { y - 1 } \right) d x + \left( y ^ { x } \ln x \right) d y$
B) $\left( x y ^ { x - 1 } \right) d x d x - \left( x ^ { y } \ln x \right) d y$
C) $\left( x y ^ { x - 1 } \right) d x + \left( x ^ { y } \ln x \right) d y$
D) $\left( y x ^ { y - 1 } \right) d x + \left( x ^ { y } \ln x \right) d y$
E) $\left( y x ^ { y - 1 } \right) d x - \left( x ^ { y } \ln x \right) d y$

Question

Let $z = x ^ { y }$ . Then the differential $d ^ { z }$ is

A) $\left( y x ^ { y - 1 } \right) d x + \left( y ^ { x } \ln x \right) d y$
B) $\left( x y ^ { x - 1 } \right) d x d x - \left( x ^ { y } \ln x \right) d y$
C) $\left( x y ^ { x - 1 } \right) d x + \left( x ^ { y } \ln x \right) d y$
D) $\left( y x ^ { y - 1 } \right) d x + \left( x ^ { y } \ln x \right) d y$
E) $\left( y x ^ { y - 1 } \right) d x - \left( x ^ { y } \ln x \right) d y$

Quizplus · Accepted Answer

The correct differential $d^z$ for $z = x^y$ is found using the chain rule and the properties of exponents. The partial derivative with respect to $x$ gives $\left( y x^{y-1} \right) dx$, and the partial derivative with respect to $y$ gives $\left( x^y \ln x \right) dy$.

Let z=xyz = x ^ { y }z=xy Then the Differential

Let $z = x ^ { y }$ Then the Differential