- #1

VikingStorm

y= [sec^2(x)] / [x^2 + 1]

dy = tanx(x^2+1) - 2x(sec^2) dx / (x^2+1)^2

That's basically what I got so far, is that it, or can I simplify more (or did I derive something wrong)?

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- Thread starter VikingStorm
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- #1

VikingStorm

y= [sec^2(x)] / [x^2 + 1]

dy = tanx(x^2+1) - 2x(sec^2) dx / (x^2+1)^2

That's basically what I got so far, is that it, or can I simplify more (or did I derive something wrong)?

- #2

- 841

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I get:

dy = 2 sec^{2}(x) [tan(x)(x^{2}+1) - x] dx / (x^{2}+1)^{2}

dy = 2 sec

- #3

VikingStorm

- #4

- 841

- 1

- #5

- 841

- 1

d(sec

Is that what you did?

- #6

VikingStorm

Originally posted by Ambitwistor

d(sec^{2}(x))/dx = tan(x)

Is that what you did?

Ah.. that must be it, I was not aware it wasn't the same backwards

*tries again*

- #7

- 841

- 1

Originally posted by VikingStorm

Ah.. that must be it, I was not aware it wasn't the same backwards

Do you mean, because d(tan(x))/dx = sec

- #8

VikingStorm

f(x)=(x+1)/sqrt(x)

f'= x^(-1/2) + (-1/2x^(-3/2))(x+1)

Where do I go from here?

f(x)=sinx + cosx, [0, 2pi]

What I did:

f'=cosx - sinx

f"=-sinx - cosx

-sinx - cosx = 0

-sinx = cosx

I'm not sure what to do next from here, divide by -sinx? and Solve cotx = 1? (And solving for x should get me my point of inflection after plugging into the original function correct? And then it would be (0,x) (x, 2pi)?)

- #9

HallsofIvy

Science Advisor

Homework Helper

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The simplest way to differentiate √(x) is to write it as

x

The simplest way to differentiate (x+1)/√(x) is to write it as f(x)=(x+1)x

Then f'(x)= (1/2)x

Where do you go from here? Since concavity depends on the second derivative, differentiate again!

f"(x)= -(1/4)x

The graph is concave upward as long as the second derivative is positive, concave downward as long as it is negative and has a point of inflection where it changes from positive to negative: you need to find where f"= 0 to divide the real line into intervals and then determine on which intervals f" is positive or negative.

"f(x)=sinx + cosx, [0, 2pi]

What I did:

f'=cosx - sinx

f"=-sinx - cosx

-sinx - cosx = 0

-sinx = cosx"

Okay, so far that's correct. Now what values of x satisfy that equation? Many people would be able to look at the equation and immediately write down the solutions (what kind of right triangle has both legs of equal length?). Yes, you can write as a "cotangent"

cot(x)= -1 (NOT 1!!) or as "tangent"- divide both sides by cos(x) to get -tan(x)= 1 or tan(x)= -1. What values of x have that property?

(If you use a calculator, I recommend you put it in "degree" mode- you should be able to recognize the "obvious" answer then.)

No, the correct answer is not of the form (x, 2pi) and certainly not (0, x).

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