Assume that all the given functions have continuous second-order partial derivatives. If z = f(x, y), where x = 9r cos(θ) and y = 9r sin(θ), find the following: A) ∂z / ∂r B) ∂z / ∂θ C) ∂^2z / ∂r ∂θ

Answer with Step-by-step explanation:We are given that all the given functions have continuous second-order partial derivatives.[tex]z=f(x,y)[/tex]Where [tex]x=9rcos\theta,y=9r sin\theta[/tex]We have to findA.[tex]\frac{\delta z}{\delta r}[/tex]We know that [tex]\frac{\delta z}{\delta r}=\frac{\delta z}{\delta x}\frac{\delta x}{\delta r}+\frac{\delta z}{\delta y}\frac{\delta y}{\delta r}[/tex]Using this formula[tex]\frac{\delta z}{\delta r}=9cos\theta \frac{\delta z}{\delta r}+9sin\theta\frac{\delta z}{\delta r}[/tex][tex]\frac{\delta z}{\delta r}=\frac{x}{r}\frac{\delta z}{\delta r}+\frac{y}{r}\frac{\delta z}\delta y}[/tex]B.[tex]\frac{\delta z}{\delta \theta}[/tex][tex]\frac{\delta z}{\delta \theta}=\frac{\delta z}{\delta x}\frac{\delta x}{\delta\theta }+\frac{\delta z}{\delta y}\frac{\delta y}{\delta\theta}[/tex][tex]\frac{\delta z}{\delta \theta}=-9rsin\theta\frac{\delta z}{\delta x}+9rcost\theta\frac{\delta z}{\delta y}[/tex][tex]\frac{\delta z}{\delta \theta}=-y\frac{\delta z}{\delta x}+x\frac{\delta z}{\delta y}[/tex]C.[tex]\frac{\delta^2 z}{\delta r\delta\theta}[/tex][tex]\frac{\delta^2 z}{\delta r\delta\theta}=-9sin\theta\frac{\delta z}{\delta x}-y\frac{\delta^2z}{\delta x^2}(9cos\theta)+9 cos\theta\frac{\delta z}{\delta y}+x\frac{\delta^2z}{\delta y^2}(9sin\theta)[/tex][tex]\frac{\delta^2 z}{\delta r\delta\theta}=-9sin\theta\frac{\delta z}{\delta x}-(81rsin\theta cos\theta)\frac{\delta^2z}{dx^2}+9cos\theta\frac{\delta z}{\delta y}+(81r cos\theta sin\theta)\frac{\delta^2z}{\delta y^2}[/tex][tex]\frac{\delta^2 z}{\delta r\delta\theta}=-\frac{y}{r}\frac{\delta z}{\delta x}-\frac{xy}{r}\frac{\delta^2}{\delta x^2}+\frac{x}{r}\frac{\delta z}{\delta y}+\frac{xy}{r}\frac{\delta^2z}{\delta y^2}[/tex]