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

[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 textbook 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

 

[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

That's what you've got to live with when you deliver too speedy answers
..

 

[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]

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. :)