this is similar to section 6.1 problem 40:\nfor ( f(x,y)=-4(2x^{2}+y^{3})^{4} ):\n(a) determine (…

Explanation:

Step1: Find the first - order partial derivative (f_x(x,y))

Using the chain rule, if (u = 2x^{2}+y^{3}), then (f(x,y)=-4u^{4}). The derivative of (f) with respect to (x) is (f_x(x,y)=-4\times4u^{3}\times4x=-64x(2x^{2}+y^{3})^{3})

Step2: Find the second - order partial derivative (f_{xx}(x,y))

Let (v=(2x^{2}+y^{3})^{3}). Then (f_x(x,y)=-64xv) Using the product rule ((uv)^\prime = u^\prime v+uv^\prime), where (u = - 64x) and (v=(2x^{2}+y^{3})^{3}) (u^\prime=-64), (v^\prime = 3(2x^{2}+y^{3})^{2}\times4x) (f_{xx}(x,y)=-64(2x^{2}+y^{3})^{3}-64x\times12x(2x^{2}+y^{3})^{2}) (=-64(2x^{2}+y^{3})^{2}(2x^{2}+y^{3}+12x^{2})) (=-64(2x^{2}+y^{3})^{2}(14x^{2}+y^{3}))

Step3: Find the first - order partial derivative (f_y(x,y))

Using the chain rule, if (u = 2x^{2}+y^{3}), then (f(x,y)=-4u^{4}) The derivative of (f) with respect to (y) is (f_y(x,y)=-4\times4u^{3}\times3y^{2}=-48y^{2}(2x^{2}+y^{3})^{3})

Step4: Find the second - order partial derivative (f_{xy}(x,y))

Let (u=-48y^{2}) and (v=(2x^{2}+y^{3})^{3}) Using the product rule ((uv)^\prime = u^\prime v+uv^\prime), where (u^\prime=-96y), (v^\prime = 3(2x^{2}+y^{3})^{2}\times4x) (f_{xy}(x,y)=-96y(2x^{2}+y^{3})^{3}-48y^{2}\times12x(2x^{2}+y^{3})^{2}) (=-96y(2x^{2}+y^{3})^{2}(2x^{2}+y^{3}+6xy))

Step5: Find the second - order partial derivative (f_{yy}(x,y))

Let (u=-48y^{2}) and (v=(2x^{2}+y^{3})^{3}) Using the product rule ((uv)^\prime = u^\prime v+uv^\prime), where (u^\prime=-96y), (v^\prime = 3(2x^{2}+y^{3})^{2}\times3y^{2}) (f_{yy}(x,y)=-96y(2x^{2}+y^{3})^{3}-48y^{2}\times9y^{2}(2x^{2}+y^{3})^{2}) (=-48(2x^{2}+y^{3})^{2}[2y(2x^{2}+y^{3}) + 9y^{4}]) (=-48(2x^{2}+y^{3})^{2}(4x^{2}y+2y^{4}+9y^{4})) (=-48(2x^{2}+y^{3})^{2}(4x^{2}y + 11y^{4}))

Answer:

(a) (-64(2x^{2}+y^{3})^{2}(14x^{2}+y^{3})) (b) (-96y(2x^{2}+y^{3})^{2}(2x^{2}+y^{3}+6xy)) (c) (-48(2x^{2}+y^{3})^{2}(4x^{2}y + 11y^{4}))