Explanation:
a. To find dy/dx when y = 2xlog e (2x), we can use the chain rule. First, we rewrite the function as y = 2x(ln(2x) + 1) where ln is the natural logarithm. Then, we can take the derivative of y with respect to x:
dy/dx = 2(ln(2x) + 1) + 2x(1/2x) = 2ln(2x) + 2 + 2/x
b. To find f′(0) when f(x) = e^x^2/e^x+1, we can take the derivative of the function with respect to x:
f′(x) = (2xe^x^2(e^x+1) - e^x^2(e^x+1)) / (e^x+1)^2
Setting x = 0, we get:
f′(0) = 2e^0(e^0 + 1) - e^0(e^0 + 1) / (e^0 + 1)^2 = 1 / (e^0 + 1)^2 = 1 / 2^2 = 1/4