Calculus problems?

Question

I'm preparing for my final examination. However, I did not come to class some day, I could not understand some problems.
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Caculate the derivatives:
d/dt intergal[from 1 to sin(t)]Cos(s^2)dx

d/dt intergal[from e^t to t^3] spr(1+x^2)dx
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The answer for (1) is Cos(Sin(t^2)(Cos(s))

How to get the right answers are very important.

Answer 1

Let's start with a simple example.

Suppose that the first one looks like:

d/dt intergal[from 1 to sin(t)]cos(x)dx

It is possible to take the integral of cos(x) by simple method. Therefore, by just taking the integral,

intergal[from 1 to sin(t)]Cos(x)dx
= [sin(x)]from 1 to sin(t)
= sin(sin(t)) - sin(1)

Note that sin(1) is a constant.
By taking its derivative respect to t,

d/dt [sin(sin(t)) - sin(1)]
= cos(sin(t))cos(t) - 0......................chain rule is used
= cos(sin(t))cos(t)

Therefore, concluding the example above,

d/dt intergal[from 1 to sin(t)]cos(x)dx = cos(sin(t))cos(t)

It resembles the second fundamental theorem of calculus, just that you have to take the derivatives of the boundary.

So, your questions can be answered as follows:

d/dt intergal[from 1 to sin(t)]Cos(x^2)dx
= Cos(Sin(t^2)(Cos(t))....................by taking the derivative of the boundary

d/dt intergal[from e^t to t^3] sqrt(1+x^2)dx
= d/dt { intergal[from e^t to 1] sqrt(1+x^2)dx + intergal[from 1 to t^3] sqrt(1+x^2)dx }

= d/dt { - intergal[from 1 to e^t] sqrt(1+x^2)dx + intergal[from 1 to t^3] sqrt(1+x^2)dx }

= d/dt { - intergal[from 1 to e^t] sqrt(1+x^2)dx } + d/dt {intergal[from 1 to t^3] sqrt(1+x^2)dx }

= -sqrt(1+e^(2t))*e^t + sqrt(1+t^6)*3t^2

Note that in order to apply second fundamental theorem of calculus, the boundary must be from some constant to a function, or simply a constant. In other words, the lower boundary must be a constant.

Answer 2

1) I am a little puzzled by the answer you give. I think you should get:

cos( (sin(t))^2 ) * cos(t)

Here's how I see it. Let f(x) = cos(x^2). Your original statement now reads:

d/dt integral[from 1 to sin(t)] f(s) ds

Let F(x) be an "antiderivative" of f(x). By that I mean that we choose F such that F'(x) = f(x). According to Newton's Fundamental Theorem of Calculus,

integral[from a to b] f = F(b) - F(a)

In this case:

d/dt integral[from 1 to sin(t)] f(s) ds = d/dt ( F(sin(t)) - F(1) )
= d/dt F(sin(t)) - d/dt F(1) = d/dt F(sin(t))

In the last equality, the last term is zero because it's the derivative of a constant. To compute what's left, we use the composite function theorem:

d/dt F(sin(t)) = f(sin(t)) * d/dt sin(t) = f(sin(t)) * cos(t)

Remember that we had defined f(x) = cos(x^2). Therefore the answer should be:

cos( (sin t)^2 ) * cos(t)


2) I don't know the symbol "spr". Are you certain you don't mean "sqr" or "sqrt"? In any case, the principle is the same. Let f(x) = spr(1+x^2), and let F be the antiderivative of f, such that F'(x) = f(x). Then:

d/dt int[from e^t to t^3] f(s) ds = d/dt ( F(t^3) - F(e^t) )
= f(t^3) * d/dt(t^3) - f(e^t) * d/dt(e^t)
= f(t^3) * 3t^2 - f(e^t) * e^t
= 3 t^2 spr(1 + t^6) - e^t spr(1 + e^(2t))

Without knowing what "spr" stands for, I cannot really simplify that result any further...

Answer 3

These problems are known as using the Fundamental Theorem of Calculus.

The key thing to note about these problems is that _absolutely NO integration is involved_. The only thing involved when solving the derivative of an integral is the chain rule.

I'm going to convert this slightly but it's the same problem.

Let
f(t) = Integral (1 to sin(t), cos(x^2)) dx

When solving for f'(t), all you would do is substitute sin(t) for every occurrance of x, use the chain rule for sin(t), and then substitute 1 for every occurrance of x, then use the chain rule on 1.

f'(t) = cos( {sin t}^2) [cos(t)] - cos ( {1}^2 ) [0]

Note the above: The { } brackets indicate that I simply replaced every occurrance of x with {sin t}, and the square brackets [ ] represent using the chain rule on the integral bounds of sin(t) and 1. The derivative of sin(t) is cos(t), and the derivative of 1 is 0.
We also do subtraction after plugging in values.

For part 2.

f(t) = Integral (e^t to t^3, sqrt(1 + x^2))dx

Again, when solving f'(t), plug in t^3 for every occurrance of x, chain the t^3, MINUS, plug in e^t for every occurrance of x, chain the e^t.

f'(x) = sqrt(1 + {t^3}^2) [3t^2] - sqrt (1 + {e^t}^2) [e^t]

As you can see: ZERO integration. This stuff covers such a small part of the course that many students either overlook it, or classes don't spend much time on it, and then when it appears on an exam, students unnecessarily lose marks for something which is so trivial to do.

I hope my explanation was clear; to recognize these problems, all you have to do is notice the problem asking you to take the derivative of an integral.

Answer 4

f (x) = x^3 - 12x + a million . . . the first spinoff set to 0 unearths turning or table sure factors f ' (x) = 3x^2 - 12 3x^2 - 12 = 0 3 * (x + 2) * (x - 2) = 0 x = 2 ... x = - 2 . . . the 2d spinoff evaluated at x = 2 and -2 determines if those factors are min, max, or neither. f ' ' (x) = 6x f ' ' (2) = 6*2 = 12 <== helpful cost exhibits x=2 is an section minimum f ' ' (-2) = 6*(-2) = -12 <== detrimental cost exhibits x=-2 is an section optimal a.) x = - 2 is a optimal, and x=2 is a minimum ... so x = - infinity to -2 is increasing x = -2 to +2 is lowering x = +2 to + infinity is increasing b.) f (-2) = (-2)^3 - 12*(-2) + a million = 17 f (2) = (2)^3 - 12*(2) + a million = - 15 c.) . . . the 2d spinoff set to 0 unearths inflection factors, or the position concavity adjustments 6x = 0 x = 0 <=== inflection factor x = - 2 is a optimal, so should be concave down concavity adjustments on the inflection factor(s) ... so x = - infinity to 0 is concave down x = 0 to + infinity is concave up

Answer 5

I am pretty good at calculus, but I am also pretty good at spelling. The word 'integral' for example has the g first.

I am not trying to be a dick here, but if you misspell the word integral, you may in fact get docked for it.

Answer 6

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Answer 7

If it wher geometry id be all over it. sorry I cant help.