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:

f'(x) = lim(h->0) [ (x+h)^(x+h) - x^x ] / h.

To solve this, we can use the limit properties and the chain rule. Let's simplify the expression:

f'(x) = lim(h->0) [ (x+h)^(x+h) - x^x ] / h
= lim(h->0) [ (x+h)^(x+h) - (x^x * (x+h)^0 ) ] / h
= lim(h->0) [ (x+h)^(x+h) - x^x * 1 ] / h.

Now, we can rewrite (x+h)^(x+h) as e^( (x+h) * ln(x+h) ), where e is the base of the natural logarithm.

f'(x) = lim(h->0) [ e^( (x+h) * ln(x+h) ) - x^x ] / h.

At this point, it is difficult to find a closed-form expression for the derivative, so we can conclude that the derivative of y = x^x cannot be easily solved using the limit definition.

b) The limit definition for the derivative of the inverse trig function from question 2, which is y = tan^(-1)(x), is given by:

f'(x) = lim(h->0) [ tan^(-1)(x + h) - tan^(-1)(x) ] / h.

To solve this, we can use the tangent addition formula:

tan(a + b) = (tan(a) + tan(b)) / (1 - tan(a) * tan(b)).

Using this formula, we can rewrite the expression as:

f'(x) = lim(h->0) [ (tan^(-1)(x) + tan^(-1)(h)) - tan^(-1)(x) ] / h
= lim(h->0) [ tan^(-1)(h) ] / h.

At this point, we can use the limit definition of the derivative of the inverse tangent function, which is:

lim(x->0) [ tan^(-1)(x) / x ] = 1.

Therefore, we can conclude that f'(x) = 1.

c) Implicit differentiation is a technique used to find the derivative of a function when it is not explicitly given in the form of y = f(x). It is particularly useful when dealing with equations where both x and y are present, such as in the examples from questions 1 and 2.

In question 1, we had the equation x^2 + y^2 = 25. By differentiating both sides of the equation with respect to x, treating y as a function of x, we can find the derivative implicitly. This allows us to find dy/dx without explicitly solving for y.

In question 2, we had the equation sin(x) + cos(y) = 1. Again, by differentiating both sides of the equation with respect to x, treating y as a function of x, we can find the derivative implicitly. This allows us to find dy/dx without explicitly solving for y.

Implicit differentiation provides a powerful tool for finding derivatives in situations where it may be difficult or impractical to solve explicitly for y in terms of x.

a) The limit definition for the derivative of y = x^x can be written as:

lim(h->0) [ ( (x+h)^(x+h) - x^x ) / h ]

To attempt to solve it, we can first simplify the expression inside the limit:
= lim(h->0) [ ( (x+h) * (x+h) * ... * (x+h) * (x+h) - x * x * ... * x * x ) / h ]
= lim(h->0) [ (x+h)^x * (x+h) - x^x ) / h ]

Next, we can expand the binomial (x+h)^x using the binomial theorem:
(x+h)^x = C(x,0) * x^0 * h^x + C(x,1) * x^1 * h^(x-1) + C(x,2) * x^2 * h^(x-2) + ...

Applying this expansion, we have:
= lim(h->0) [ ( C(x,0) * x^0 * h^x * (x+h) + C(x,1) * x^1 * h^(x-1) * (x+h) + ... ) - x^x ) / h ]
= lim(h->0) [ ( C(x,0) * x^0 * h^x * x + C(x,1) * x^1 * h^(x-1) * x + ... + C(x,0) * x^0 * h^x * h + C(x,1) * x^1 * h^(x-1) * h + ... ) - x^x ) / h ]

Now, we can simplify further by canceling out x^x terms and factoring out h from each term inside the limit:
= lim(h->0) [ ( C(x,0) * x * h^x * x + C(x,1) * x * h^(x-1) * x + ... + C(x,0) * h^x * h + C(x,1) * x * h^(x-1) * h + ... ) / h ]
= lim(h->0) [ ( x*C(x,0)*h^x + x*C(x,1)*h^(x-1) + ... + C(x,0)*h^(x+1) + C(x,1)*x*h^x + ... ) / h ]

Now, to evaluate this limit, we can divide each term by h and then take the limit as h approaches 0:
= lim(h->0) [ x*C(x,0)*h^(x-1) + x*C(x,1)*h^(x-2) + ... + (C(x,0)*h^x + C(x,1)*x*h^(x-1) + ...) ]

At this point, simplification may depend on the specific value of x. Therefore, to fully solve the limit definition, we would need to substitute a specific value for x and evaluate the limit using the simplification techniques appropriate for that value.

b) The limit definition for the derivative of the inverse trig function from question 2 can be written as:
lim(h->0) [ ( arccos(x+h) - arccos(x) ) / h ]

To attempt to solve it, we can use the properties of inverse trigonometric functions and trigonometric identities to manipulate the expression. However, this process could become quite involved and may rely on specific properties of the inverse trig function used in question 2. Therefore, the specific solution would require knowing the specific inverse trig function involved.

c) Implicit differentiation allows us to find the derivative of a function when it is not explicitly expressed in terms of one variable. It is particularly useful when we have equations in which it is not easy to solve for one variable explicitly.

For example, in question 1, we had the equation x^2 + y^3 = 10. If we try to solve this equation for y explicitly, it becomes complicated as we have a power of 3. However, by differentiating implicitly with respect to x, we can find the derivative of y with respect to x without explicitly solving for y.

Using the chain rule, we differentiate both sides of the equation with respect to x:
(d/dx) [x^2 + y^3] = (d/dx) [10]
2x + 3y^2 * (dy/dx) = 0

Here, dy/dx represents the derivative of y with respect to x. We can solve this equation for dy/dx to find the derivative of y with respect to x even without explicitly knowing y as a function of x.

Similarly, in question 2, we had the equation sin(x) + cos(y) = 1. By differentiating implicitly with respect to x, we can find the derivative of y with respect to x, even without solving for y explicitly.

Implicit differentiation is a powerful technique in calculus that allows us to find derivatives in situations where explicit solutions are not readily available. It enables us to work with equations involving multiple variables and complicated functions, providing a flexible tool for finding derivatives.