Answer to Question #339770 in Calculus for BOB

Question

Suppose z is a function of x and y, and tan√y²+x² = z^xe^6y. Solve for ∂z/∂x and ∂z/∂y.

Answer 1

$$\sqrt{y^2+x^2}=z^xe^{6y}$$

$$\dfrac{1}{2}\ln(y^2+x^2)=x\ln z+6y$$

Differentiate both sides with respect to $x$

$$\dfrac{2x}{2(y^2+x^2)}=\ln z+\dfrac{x}{z}(\dfrac{\partial z}{\partial x})$$

$$\dfrac{\partial z}{\partial x}=\dfrac{z}{y^2+x^2}-\dfrac{z\ln z}{x}$$

$$\dfrac{1}{2}\ln(y^2+x^2)=x\ln z+6y$$

Differentiate both sides with respect to $y$

$$\dfrac{2y}{2(y^2+x^2)}=\dfrac{x}{z}(\dfrac{\partial z}{\partial y})+6$$

$$\dfrac{\partial z}{\partial y}=\dfrac{yz}{x(y^2+x^2)}-\dfrac{6z}{x}$$