answer:The divergence of the vector field vec{F} is given by the sum of the partial derivatives of its components with respect to their respective variables. Applying the divergence theorem: text{div}(vec{F}) = frac{partial f}{partial x} + frac{partial g}{partial y} + frac{partial h}{partial z} Substituting the given functions: text{div}(vec{F}) = frac{partial (x^4)}{partial x} + frac{partial (cos(y))}{partial y} + frac{partial (tanh(z))}{partial z} text{div}(vec{F}) = 4x^3 - sin(y) + text{sech}^2(z) Therefore, the divergence of the vector field is 4x^3 - sin(y) + text{sech}^2(z).

answer:The glomerulus and the glomerular capsule form the filtration membrane in the nephron. The glomerulus is a network of tiny blood vessels, and the glomerular capsule is a cup-shaped structure that surrounds the glomerulus. Together, these two structures form the filtration membrane, which allows water and small molecules to pass through while preventing larger molecules, such as proteins, from passing through.

answer:The product D v is computed by multiplying the rows of matrix D with the columns of vector v. The resulting matrix will have dimensions (2times1). D v = left[begin{array}{ccc} -2 & 5 & 4 5 & 1 & 5 end{array}right] left[begin{array}{c} 0 0 -2 end{array}right] = left[begin{array}{c} ? ? end{array}right] To find the entries of the resulting vector, multiply the corresponding elements from each row of D with the elements of v and sum the products: First entry: (-2 cdot 0) + (5 cdot 0) + (4 cdot -2) = -8 Second entry: (5 cdot 0) + (1 cdot 0) + (5 cdot -2) = -10 So, the product D v is: left[begin{array}{c} -8 -10 end{array}right]

answer:{eq}eqalign{ & {text{If }},f,{text{ is a continuous and integrable function in an interval }}left[ {a,b} right]{text{, and }},y = Fleft( x right),{text{ is defined by }} cr & {text{the following integral: }},y = Fleft( x right) = intlimits_a^{gleft( x right)} {fleft( t right)} dt{text{. Then the derivative of the function }},y = Fleft( x right){text{ }} cr & {text{using the Fundamental Theorem of Calculus is calculated as follows:}} cr & ,,,,{y^prime } = {F^prime }left( x right) = frac{d}{{dx}}left( {intlimits_a^{gleft( x right)} {fleft( t right)} dt} right) = fleft( {gleft( x right)} right) cdot {g^prime }left( x right) cr & {text{We have in this particular case that the function }},y{text{ = }}F(x),{text{ is defined by the following integral:}} cr & ,,,,y = Fleft( x right) = int_0^{tan x} {sqrt t ,dt} cr & {text{Then}}{text{, we have }},fleft( t right) = sqrt t ,{text{ is a continuous function and integrable for all }},t geqslant 0,{text{ and }} cr & gleft( x right) = tan x{text{. So}}{text{, applying the Fundamental Theorem we have}}: cr & ,,,,,,,,,{y^prime } = {F^prime }left( x right) = frac{d}{{dx}}left( {int_0^{tan x} {sqrt t ,dt} } right) cr & ,,,,,,,,,{y^prime } = fleft( {gleft( x right)} right) cdot {g^prime }left( x right) = sqrt {tan x} cdot {left( {tan x} right)^prime } cr & ,,,,,,,,,{y^prime } = {F^prime }left( x right) = sqrt {tan x} cdot {sec ^2}x cr & {text{Therefore}}{text{, }}boxed{{y^prime } = frac{d}{{dx}}left( {int_0^{tan x} {sqrt t ,dt} } right) = {{sec }^2}x cdot sqrt {tan x} } cr} {/eq}