Calculus help?

Question

..................x^2
g(x)=.....⌠ dt
..............-----------
..............⌡√(4+t^4)
.................tan(x)

yeah sorry i meant that i needed g ' (x)

Answer 1

I assume that you want to solve g'(x) for

g(x) = Integral ( (1/sqrt(4-t^4)) dt from tan(x) to x^2.

This is known as the Fundamental Theorem of Calculus, because it involves taking the derivative of an integral. When solving g'(x), know that ABSOLUTELY NO DIFFERENTIATION OR INTEGRATION IS INVOLVED! The only thing that is involved is the chain rule.

To solve this type of problem, all you have to do is replace t with the top function, MINUS replacing t with the bottom function, WHILE applying the chain rule.

g(x) = Integral ( (1/sqrt(4-t^4)) dt from tan(x) to x^2.

g'(x) = [ 1/sqrt(4 - (x^2)^4) ] * (2x) - [1/sqrt(4 - [(tanx)]^4)] * [sec(x)]^2

That's all there is to these questions! Notice the 2x and [sec(x)]^2. Those are the derivatives of x^2 and tan(x) respectively.

As you can see, all you did was replace the t with the upper and lower functions, and applied the chain rule to each.
These types of questions cover such a small part of Calculus that they are often overlooked (and the textbook usually doesn't cover these in detail), which is why I'm telling you right now, that this is the proper method to do it.

ZERO differentiation. ZERO integration. Just plug in x^2 for t, apply the chain rule, plug in tan(x) for t, apply the chain rule.

Please try and take this knowledge with you when a student comes up to you asks you the exact same question.

Answer 2

First of all, I'm assuming that you meant to put an apostrophe in the g(x), because it looks like you're solving for the integral, not the derivtive.
For the first integral, you add 1 to the power, so it would be x^3 and then think - if you were to take derivative of x^3, it would be 3x^2, but the derivative given to you is only of 1x^2. Therefore, if you multiply 1/3 by x^3, the derivative of that is x^2. Your answer then is g(x) = 1/3x^3.
The second integral is a little harder. Having a square right sign over (4+t^4) is the same as putting a power of half over that expression like this: (4+t^4)^1/2. Hmm... actually, I'm only sure of that one until that step. Sorry, hope that still helped you some!