How Accurate Is My Calculus Homework Solution?

[VikingStorm]

I'm having problems with this one:

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)?

 

[VikingStorm]

Would you be kind enough to explain parts of that, I don't seem to be able to see how you got the top.

 

[VikingStorm]

Originally posted by Ambitwistor
I can arrive at your expression if I were to incorrectly use,

d(sec²(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*

 

[VikingStorm]

I'm having a bit of trouble with these as well (finding points of inflection, and concavity):

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)?)

 

[HallsofIvy]

With as much misunderstanding as I see here, you need to go to your teacher for assistance. (General rule: NEVER try to fool your teacher into thinking you can do the homework!)

The simplest way to differentiate √(x) is to write it as
x^1/2.

The simplest way to differentiate (x+1)/√(x) is to write it as f(x)=(x+1)x^-1/2= x^1/2+ x^-1/2.

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

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

f"(x)= -(1/4)x^-3/2- (3/4)x^-5/2.

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).