• June 22nd 2011, 06:20 PM
sara213
I have this problem that I'm going to make the derivative of:

$6000/(1+49(0.6)^t$

The next step i'm confused about:

$6000/(1+49(0.6)^t)^2 d/dx (1+49(0.6)^t)$

I think I am overlooking something here...

How does the $(1+49(0.6)^t)^2$come to be squared? Underneath the 6000.
• June 22nd 2011, 06:43 PM
TKHunny
Standard Exponent Rule, isn't it?

For suitable values of 'a', we have $\frac{d}{dx}\;x^{a} = a\cdot x^{a-1}$

a = -1 is a suitable value.
• June 22nd 2011, 07:59 PM
Soroban
Hello, sara213!

Quote:

Differentiate: . $f(x) \:=\:\frac{6000}{1+49(0.6^x)}$

The next step i'm confused about:

$f'(x) \;=\; \frac{6000}{\left[1+49(0.6^x)\right]^2}\,\frac{d}{dx}\left[1 + 49(0.6^x)\right]$ . not quite right

I think I am overlooking something here . . .

How does the $(1+49\!\cdot\!0.6^t)^2$come to be squared? Underneath the 6000?

Are you sure you know your differentiation rules?

You can use the Quotient Rule,
. . which ends with "over the denominator squared".

Or use an easier way . . .

We have: . $f(x) \:=\:6000\left[1 + 49(0.6^x)\right]^{-1}$

Chain Rule: . $f'(x) \;=\;6000\cdot(\text{-}1)\left[1 + 49(0.6^x)\right]^{-2} \cdot \frac{d}{dx}\left[1 + 49(0.6^x)\right]$

. . . . . . . . . . $f'(x) \;=\;\frac{-6000}{[1 + 49(0.6^x)]^2} \cdot \frac{d}{dx}\left[1 + 49(0.6^x)\right]$

• June 22nd 2011, 08:09 PM
sara213
Quote:

Originally Posted by Soroban
Hello, sara213!

Are you sure you know your differentiation rules?

You can use the Quotient Rule,
. . which ends with "over the denominator squared".

Or use an easier way . . .

We have: . $f(x) \:=\:6000\left[1 + 49(0.6^x)\right]^{-1}$

Chain Rule: . $f'(x) \;=\;6000\cdot(\text{-}1)\left[1 + 49(0.6^x)\right]^{-2} \cdot \frac{d}{dx}\left[1 + 49(0.6^x)\right]$

. . . . . . . . . . $f'(x) \;=\;\frac{-6000}{[1 + 49(0.6^x)]^2} \cdot \frac{d}{dx}\left[1 + 49(0.6^x)\right]$

I thought i did but I think the notation : d/dx is throwing me off. I like f(x) and g(x) stuff for example:

$\;=\;\frac{g(x) * f'(x) - f(x) * g'(x)}{g(x)^2}$
• June 23rd 2011, 12:13 AM
bugatti79
Quote:

Originally Posted by sara213
I thought i did but I think the notation : d/dx is throwing me off. I like f(x) and g(x) stuff for example:

$\;=\;\frac{g(x) * f'(x) - f(x) * g'(x)}{g(x)^2}$

If you want to use your notation then

$h(t) =\frac{6000}{1+49(0.6^t))}$

let $f(t)=6000$ and $g(t)=(1+46(0.6^t))$

$h'(t)=\frac{-6000* \frac{d}{dt} (1+49(0.6^t))}{(1+49(0.6^t))^2}$
• June 23rd 2011, 04:15 AM
HallsofIvy
When your numerator is a constant, as here, it is perhaps easier to think of the problem as having a negative exponent:
$\frac{6000}{1+ 49(.06)^t}= 6000(1+ 49(.06)^t)^{-1}$

Now, that the derivative is $6000(-1)(1+ 49(.06)^t)^{-2}(d/dx(1+ 49(.06)^t)$

So you can think of the denominator being $(1+ 49(.06)^t)$ being squared as a result of either the quotient rule or simply the power rule: $(x^n)'= nx^{n-1}$ with n= -1 so that n-1= -1-1= -2.

Note that to finish you will need the fact that the derivative of $a^t$ is $a^t(ln(a))$.
• June 23rd 2011, 04:30 AM
Plato
Quote:

Originally Posted by sara213
I have this problem that I'm going to make the derivative of:
$6000/(1+49(0.6)^{\color{red}t}$

The next step i'm confused about:
$6000/(1+49(0.6)^t)^2 d/\color{red}dx (1+49(0.6)^t)$

$\text{Is the variable }t\text{ or }x~?$
• June 23rd 2011, 05:06 AM
sara213
$\text{Is the variable }t\text{ or }x~?$