Question #339770

Suppose z is a function of x and y, and tan√y2+x2 = zxe6y. Solve for ∂z/∂x and ∂z/∂y.


Expert's answer

y2+x2=zxe6y\sqrt{y^2+x^2}=z^xe^{6y}

12ln⁡(y2+x2)=xln⁡z+6y\dfrac{1}{2}\ln(y^2+x^2)=x\ln z+6y

Differentiate both sides with respect to xx


2x2(y2+x2)=ln⁡z+xz(∂z∂x)\dfrac{2x}{2(y^2+x^2)}=\ln z+\dfrac{x}{z}(\dfrac{\partial z}{\partial x})


∂z∂x=zy2+x2−zln⁡zx\dfrac{\partial z}{\partial x}=\dfrac{z}{y^2+x^2}-\dfrac{z\ln z}{x}



12ln⁡(y2+x2)=xln⁡z+6y\dfrac{1}{2}\ln(y^2+x^2)=x\ln z+6y

Differentiate both sides with respect to yy


2y2(y2+x2)=xz(∂z∂y)+6\dfrac{2y}{2(y^2+x^2)}=\dfrac{x}{z}(\dfrac{\partial z}{\partial y})+6

∂z∂y=yzx(y2+x2)−6zx\dfrac{\partial z}{\partial y}=\dfrac{yz}{x(y^2+x^2)}-\dfrac{6z}{x}


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