Suppose that z = f(x,y) is a function with partial derivatives partial f/ partial x = xy, partial...

Question:

Suppose that {eq}z = f(x,y) {/eq} is a function with partial derivatives {eq}\dfrac{\partial f}{\partial x}=xy,\dfrac{\partial f}{\partial y}=y {/eq} where {eq}x = x(u, v) = 2u {/eq} and {eq}y = y(u,v) = uv {/eq}. Then, {eq}\dfrac{\partial z}{\partial v} {/eq}.

Partial derivative:

Let us consider a function {eq}z = f\left( {x,y} \right){/eq} , where {eq}x\left( s \right) = y\left( s \right) = s{/eq} , then the partial derivative of this function {eq}z = f\left( {x,y} \right){/eq} with respect to {eq}s{/eq} is given by using chain rule as {eq}\dfrac{{\partial z}}{{\partial s}} = \dfrac{{\partial f}}{{\partial x}}\dfrac{{\partial x}}{{\partial s}} + \dfrac{{\partial f}}{{\partial y}}\dfrac{{\partial y}}{{\partial s}}{/eq} . Here, {eq}\dfrac{{\partial f}}{{\partial x}}{/eq} , {eq}\dfrac{{\partial f}}{{\partial y}}{/eq} are the partial derivatives of {eq}f{/eq} with respect to {eq}x{/eq} , {eq}y{/eq} respectively.