How can you use K to determine is a reaction is product favored or reactant favored?
K is a ratio of the concentration of products to the concentration of reactants. So, if K is a lot bigger than 1, then then the concentration of products is much higher than the concentration of reactants and the reaction is product favored. On the other hand, if K is a lot smaller than 1, then the concentration of reactants is much higher than the concentration of products, and the reaction is reactant favored.
What is the difference between direct immunofluorescence and indirect immunofluorescence, and why would you use one over the other?
Direct immunofluorescence uses only a primary antibody conjugated to a fluorophore that recognize and binds to an antigen on a cell. Indirect immunofluorescence uses an unconjugated primary antibody to bind to an antigen on a cell, and then a secondary antibody conjugated to a fluorophore that binds to the primary antibody. Direct immunofluorescence is often more expensive than indirect, and does not have as wide of a range of chemistries, but it is often more specific and has a simpler protocol than indirect. Besides the lower cost and wider range of chemistries, indirect can be more advantageous than direct because it can have a stronger signal since multiple secondary antibodies can bind to one primary antibody.
Find the derivative with respect to y of the equation f(x,y) = (3x)*(y^4) + (x^3)*(y^2).
When you take the derivative of a multivariable function, you treat all of the components of the equation as a constant, except the variable you are taking the derivative with respect to. For this problem, this means that you treat both the 3x in the first term of the equation and the x^3 in the second term of the equation as a constant. You can think of them as C1 and C2, so you have C1*y^4 + C2*y^2. To take the derivative of this equation, put the exponent from the y^4 in front of the first term and subtract 1 from the exponent, so you have 4*C1*y^3. Then repeat the same process with the second term, so you have 2*C2*y. Then you can plug 3x back in for C1 and x^3 back in for C2, ending up with a final answer of 4*(3x)*(y^3) + 2*(x^3)*(y) or (12x)*(y^3) + 2*(x^3)*(y).