derivaties

[UrbanXrisis]

y=tanx
y'=sec^2x

does y' also equal to (secx)^2 ?

[arildno]

sec^{2}x\equiv(sec{x})^{2}
The first is simply a fancy way of writing the square of secx

[Diane_]

Yes. sec^2x is just the standard way of writing (sec x)^2. It helps avoid confusion with sec (x^2).

[UrbanXrisis]

my text book also has something about
d/dx a^u = a^u ln a du/dx

what's that about?

[arildno]

That's a chain rule use, with "a" a constant, "u" a function of "x".

[Pyrrhus]

When you have something like

y = a^u

you can solve it as

ln y = ln a^u

ln y = u ln a

Derivating the above we will have

\frac{y'}{y} = u' ln a

y' = y u' ln a

y' = a^u ln a u'

[UrbanXrisis]

I thought chain rule was something like:
d/dx f(q(x))=f'(g(x))*q'(x)

[Diane_]

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. :)

[arildno]

That's right:
\frac{d}{du}a^{u}=a^{u}ln(a)
Hence,
\frac{d}{dx}a^{u(x)}=(\frac{d}{du}a^{u})\frac{du}{ dx}=a^{u(x)}ln(a)\frac{du}{dx}

[Pyrrhus]

Bah, you both did it more fancy :cry:

[UrbanXrisis]

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
???

[arildno]

Bah, you both did it more fancy :cry:
That's what you've gotta live with when you deliver too speedy answers
..:biggrin:

[arildno]

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???

[UrbanXrisis]

: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?

[arildno]

I'm not following..used???

[UrbanXrisis]

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

[Pyrrhus]

He means a function, like f(x) = 3^x

[UrbanXrisis]

ohhhh, I understand totally what it is used for now.... the "a" has to be a number without a variable!

[HallsofIvy]

f(x)= 2^{x}

\frac{df}{dx}= (ln 2)2^x

More generally,

f(x)= 3^{x^2- 3x+ 2}

\frac{df}{dx}= (2x- 3)(ln 3)3^{x^2- 3x+ 2}

[Diane_]

I'm not following..used???

Here's to pure mathematics - may it never be of any use to anyone. :)