Second derivative

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Homework Statement


Find the second derivative of 3sec[tex]\sqrt{x}[/tex]


Homework Equations


derivative of sec u = secu(tanu)*u'

product rule- (F*G)'=f'*g+g'*f
quotient rule- (F/G)'=(f'*g-f*g')/g^2

The Attempt at a Solution


So i did the first derivative of it, getting 3sin[tex]\sqrt{x}[/tex]/{2[tex]\sqrt{x}[/tex]*cos^2[tex]\sqrt{x}[/tex]}

After this point, I started to simply take this derivative again. However, my paper says to find d^2y/d^2x, which is confusing to me since there is no y term in this, so is it just simply taking the derivative twice?
 

Answers and Replies

  • #2
Dick
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Yes, it's simply taking the derivative twice. It's d/dx(dy/dx).
 
  • #3
I would say yes .. take the derivative again .. I think d^2y/d^2x, should read d2y/dx2 ...

I have the first derivative to be f' = (3/2) ((sec[tex]\sqrt{}x[/tex])[tex]\cdot[/tex](tan[tex]\sqrt{}x[/tex]))/[tex]\sqrt{}x[/tex]
 
  • #4
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Danger,
If you were to simplify that to just sin and cos, then it would be the same as mine. I did this so the derivatives of the trig functions in the quotient rule portion would be easier (just -sin or cos).
 
  • #5
try this one
substitute for x^1/2

let y = 3sec(x^1/2)
let u = x^1/2

thus
y = 3sec(u)
y' = 3sec(u)tan(u)
now second derivative is
y'' = 3[sec(u)tan(u).tan(u) + sec^2(u)sec(u)]
y'' = 3[sec(u)tan^2(u) + sec^3(u)]

substuting back for u

y'' = 3[sec(x^1/2)tan^2(x^1/2) + sec^3(x^1/2)]

any comments???
 
Last edited:
  • #6
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whoaaaaaaa,
Electrophysics, I am not that far into derivitives to know whether or not that is an application, but are you saying I can do that? or are you like saying, maybe you could do that. That seems so much easier than what i did (which took about a page and a half of paper).
 
  • #7
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try this one
substitute for x^1/2

let y = 3sec(x^1/2)
let u = x^1/2

thus
y = 3sec(u)
y' = 3sec(u)tan(u)
now second derivative is
y'' = 3[sec(u)tan(u).tan(u) + sec^2(u)sec(u)]
y'' = 3[sec(u)tan^2(u) + sec^3(u)]

substuting back for u

y'' = 3[sec(x^1/2)tan^2(x^1/2) + sec^3(x^1/2)]

any comments???
And, wouldn't the first derivative of 3secu be
3sec(u)tan(u)(u')??? often it doens't matter because it is just an x or whatever, but since squareroot of x would have a derivative, doens't that come into play.?
 
  • #8
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Electro physics - u have have calculated d^2y/du^2
to calculate d^2y/dx^2 you must multiply your answer by d^2u/dx^2
ie: by the second derivative of x^1/2.

This is a simple chain rule application.
 
  • #9
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hesbon,
If i take the answer that electro got and multiply it by f'' of x^1/2 i'll get the correct answer?


Also, if i were to take the first derivative, and simpify it so that it is in sin and cos, i will still get the same derivative as if i keep it in secants and tangents correct?
 
  • #10
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U-substitution just makes it more complicated.

Just take the first derivative.

that would be (1/2)*3 * sec(sqrt[x]) * tan(sqrt[x])

So you have sec * tan * 3/2

Take the derivative again using the chain rule.
 
  • #11
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And, wouldn't the first derivative of 3secu be
3sec(u)tan(u)(u')??? often it doens't matter because it is just an x or whatever, but since squareroot of x would have a derivative, doens't that come into play.?
Yes and yes.
The work posted by ElectroPhysics is wrong if by y' he means dy/dx, and by y'' he means d^2y/dx^2.

If both of these derivatives are with respect to u, then the work is right, but incomplete, as the original problem asked for the first and second derivatives w.r.t. x.
let u = x^1/2

thus
y = 3sec(u)
y' = 3sec(u)tan(u)
now second derivative is
y'' = 3[sec(u)tan(u).tan(u) + sec^2(u)sec(u)]
y'' = 3[sec(u)tan^2(u) + sec^3(u)]

substuting back for u

y'' = 3[sec(x^1/2)tan^2(x^1/2) + sec^3(x^1/2)]
The prime notation is fine as long as it's understood with respect to which variable differentiation occurs. As soon as you introduce a third variable, there's the potential for losing clarity.

Here's the first derivative again, this time in a more explicit form:
dy/du = 3sec(u)tan(u)
dy/dx = dy/du * du/dx = 3sec(sqrt(x))tan(sqrt(x)) * d(sqrt(x))/dx
 
  • #12
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assuming that he has calculated the derivative with respect to u correctly than ..yes all u have to do is multiple by second derivative of x^1/2.

you can present the initial function in many equivalent trignometric forms
The derivatives of all of them will also be equivalent.
however they may appear to be different, but with a little trig manipulation you will find that they are infact identical.
so the answer to your second question is yes.
 
  • #13
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thank you heshbon!


I dont know how long it would take you/if it is alot to ask, but i was wondering if you could post what you would get if you were to take the derivative of

3sin[tex]\sqrt{x}[/tex]/(2[tex]\sqrt{x}[/tex]cos^2[tex]\sqrt{x}[/tex])

This is the first derivative of 3sec[tex]\sqrt{x}[/tex] after simplifying to sin and cos.

-3(sin[tex]\sqrt{x}[/tex]/(4x^3/2cos^2[tex]\sqrt{x}[/tex])+(sin^2[tex]\sqrt{x}[/tex]cos[tex]\sqrt{x}[/tex])/(2x(cos^2[tex]\sqrt{x}[/tex])^2)+cos[tex]\sqrt{x}[/tex]/(4xcos^2[tex]\sqrt{x}[/tex])

wow, that is just about impossible to read, but does that look like it would work?
 

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