Y=tanx y'=sec^2x does y' also equal to (secx)^2 ?

  • #1
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y=tanx
y'=sec^2x

does y' also equal to (secx)^2 ?
 
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  • #2
arildno
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[tex]sec^{2}x\equiv(sec{x})^{2}[/tex]
The first is simply a fancy way of writing the square of secx
 
  • #3
Diane_
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Yes. sec^2x is just the standard way of writing (sec x)^2. It helps avoid confusion with sec (x^2).
 
  • #4
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my text book also has something about
d/dx a^u = a^u ln a du/dx

what's that about?
 
  • #5
arildno
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That's a chain rule use, with "a" a constant, "u" a function of "x".
 
  • #6
Pyrrhus
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When you have something like

[tex] y = a^u [/tex]

you can solve it as

[tex] ln y = ln a^u [/tex]

[tex] ln y = u ln a [/tex]

Derivating the above we will have

[tex] \frac{y'}{y} = u' ln a [/tex]

[tex] y' = y u' ln a [/tex]

[tex] y' = a^u ln a u'[/tex]
 
  • #7
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I thought chain rule was something like:
d/dx f(q(x))=f'(g(x))*q'(x)
 
  • #8
Diane_
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It can be derived using the fundamental rules.

Let a^u = y

ln (a^u) = ln(y)

u ln(a) = ln(y)

d/dx(u ln(a)) = d/dx (ln(y))

For the first part, remember the product rule:

d/dx(u ln(a)) = u d/dx(ln(a)) + ln(a) d/dx(u)

Since a is a constant, d/dx(ln(a)) = 0, so

d/dx(u ln(a)) = ln(a) du/dx

For the other side, use the chain rule:

d/dx (ln(y)) = 1/y dy/dx

So:

ln(a) du/dx = 1/y dy/dx

dy/dx = y ln(a) du/dx

Remember that y = a^u, so

d/dx(a^u) = a^u ln(a) du/dx

QED = quod erat demonstrandum = quite easily done. :)
 
  • #9
arildno
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That's right:
[tex]\frac{d}{du}a^{u}=a^{u}ln(a)[/tex]
Hence,
[tex]\frac{d}{dx}a^{u(x)}=(\frac{d}{du}a^{u})\frac{du}{dx}=a^{u(x)}ln(a)\frac{du}{dx}[/tex]
 
  • #10
Pyrrhus
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Bah, you both did it more fancy :cry:
 
  • #11
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so

d/dx (a^2+2)^2=2(a^2+2)*2a
but it also equals
d/dx (a^2+2)^2=(a^2+2)^2*ln(a^2+2)*2a
???
 
  • #12
arildno
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Cyclovenom said:
Bah, you both did it more fancy :cry:
That's what you've gotta live with when you deliver too speedy answers
..:biggrin:
 
  • #13
arildno
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UrbanXrisis said:
so

d/dx (a^2+2)^2=2(a^2+2)*2a
but it also equals
d/dx (a^2+2)^2=(a^2+2)^2*ln(a^2+2)*2a
???
WHAT?????????
2 is not a varying function, is it?
Besides, shouldn't it be d/da???
 
  • #14
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:rofl:

I guess I was confused. Can you give an example of where the derivative of a^u can be used? As in show me an easy example problem?
 
  • #15
arildno
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I'm not following..used???
 
  • #16
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As in do a sample problem, like how I made up the equation (a^2+2)^2 and found the derivative with the chain rule: d/dx (a^2+2)^2=2(a^2+2)*2a
 
  • #17
Pyrrhus
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He means a function, like [tex] f(x) = 3^x [/tex]
 
  • #18
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ohhhh, I understand totally what it is used for now.... the "a" has to be a number without a variable!
 
  • #19
HallsofIvy
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[tex] f(x)= 2^{x}[/tex]

[tex] \frac{df}{dx}= (ln 2)2^x [/tex]

More generally,

[tex] f(x)= 3^{x^2- 3x+ 2}[/tex]

[tex]\frac{df}{dx}= (2x- 3)(ln 3)3^{x^2- 3x+ 2}[/tex]
 
  • #20
Diane_
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arildno said:
I'm not following..used???
Here's to pure mathematics - may it never be of any use to anyone. :)
 

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