3. a) Write out the limit definition for the derivative of y = xx. Attempt to solve it.

b) Write out the limit definition for the derivative of the inverse trig function from question 2. Attempt to solve it.
c) Discuss the value of implicit differentiation. (Use questions 1 and 2 as examples.)

a) The limit definition for the derivative of y = x^x is given by:

lim (h→0) [(x + h)^(x + h) - x^x] / h

To solve it, we can simplify the expression using some properties of exponents:

lim (h→0) [(x + h)^(x + h) - x^x] / h
= lim (h→0) [(x + h)^(x + h) / x^x - 1] / h

Next, we can use the fact that (a^b)^c = a^(b*c) to rewrite the numerator:

lim (h→0) [((x + h) / x)^((x + h)*x) - 1] / h

Now, we can use the fact that e^ln(a) = a to rewrite the exponent:

lim (h→0) [e^((x + h)*x*ln((x + h) / x)) - 1] / h

At this point, we can simplify further by using Taylor series expansion for e^x:

lim (h→0) [(1 + (x + h)*x*ln((x + h) / x) + O(h^2)) - 1] / h
= lim (h→0) [(x + h)*x*ln((x + h) / x) + O(h^2)] / h

By canceling out the h in the numerator and denominator, we get:

lim (h→0) (x*x*ln((x + h) / x) + O(h^2))
= x^x * ln((x + h) / x)

So, the derivative of y = x^x is x^x * ln((x + h) / x).

b) The limit definition for the derivative of the inverse trig function from question 2, let's assume it's y = arcsin(x), is given by:

lim (h→0) [arcsin(x + h) - arcsin(x)] / h

To solve it, we need to apply a trigonometric identity called the Pythagorean identity, which states that sin^2(x) + cos^2(x) = 1. Through this identity, we get:

lim (h→0) [sin(arcsin(x + h)) - sin(arcsin(x))] / h

Using the inverse trig function property, sin(arcsin(z)) = z:

lim (h→0) [(x + h) - x] / h
= lim (h→0) h / h
= 1

So, the derivative of y = arcsin(x) is 1.

c) Implicit differentiation is a technique used when the dependent and independent variables are not explicitly written as functions, but are instead related through an equation. It allows us to find the derivative of the dependent variable with respect to the independent variable, even if the equation is not explicitly solved for the dependent variable.

In question 1, implicit differentiation was used to find the derivative of y with respect to x. By differentiating both sides of the equation with respect to x and treating y as an implicit function of x, we were able to find the derivative dy/dx.

In question 2, implicit differentiation was used to find the derivative of arcsin(x) with respect to x. Since arcsin(x) is the inverse function of sin(x), and the inverse trig functions are defined implicitly, we needed to use implicit differentiation to find the derivative.

Implicit differentiation can be a powerful tool in situations where it's difficult or impossible to solve the equation explicitly for the dependent variable. It allows us to find derivatives even when functions are defined implicitly and not in a explicitly solved form.