# Derivative help

• July 25th 2007, 06:47 PM
starswept
Derivative help
Hi, I have two questions with derivatives that I'm struggling with:

1) If f(x)= (xlnx)^2, find all points at which the graph of f(x) has a horizontal tangent line.

I really have no idea where to start with this one- how do you find the derivative when the entire ln, not just the function, is squared?

2) Formula for pH is: pH = -ln (H) divided by ln (10).

Ingredients are being added to tomatoes so that the concentration of hydrogen ions in the mixture is given by H(t) = 30-5t-25 (e^(-1/5) - 1) moles per litre. Determine the rate of change of the pH value with respect to time after 10 s.

I'd greatly appreciate any help. Thank you!
• July 25th 2007, 06:49 PM
tukeywilliams
So $f(x) = (x \ln x)^2$. Then we want to set $f'(x) = 0$.

So $f'(x) = 2x \ln x \left(1 + \ln x \right)$ or $0 =2x \ln x \left(1 + \ln x \right)$. So $x = 0,1$.

$\text{pH} = (\frac{1}{\ln 10}-\ln(30-5t-25(e^{-1/5}-1)))$.

Then $\frac{d\text{PH}}{dt} = \frac{1}{\ln 10}\left(-\frac{1}{30-5t}(5) \right)$. So at $t = 10$, $\frac{d\text{PH}}{dt} = \frac{1}{4 \ln 10}$.
• July 26th 2007, 05:30 AM
topsquark
Quote:

Originally Posted by tukeywilliams
So $f(x) = (x \ln x)^2$. Then we want to set $f'(x) = 0$.

So $f'(x) = 2x \ln x \left(1 + \ln x \right)$ or $0 =2x \ln x \left(1 + \ln x \right)$. So $x = 0,1$.

There's a solution missing from the above solved equation:
$0 =2x \ln x \left(1 + \ln x \right)$

Thus
$2x = 0$ ==> $x = 0$
or
$ln(x) = 0$ ==> $x = 1$
or
$1 + ln(x) = 0$ ==> $x = \frac{1}{e}$

Quote:

Originally Posted by tukeywilliams
$\text{pH} = (\frac{1}{\ln 10}-\ln(30-5t-25(e^{-1/5}-1)))$.

Then $\frac{d\text{PH}}{dt} = \frac{1}{\ln 10}\left(-\frac{1}{30-5t}(5) \right)$. So at $t = 10$, $\frac{d\text{PH}}{dt} = \frac{1}{4 \ln 10}$.

There's a slight typo in your pH formula and I'm not sure what happened when you took the derivative:
$pH = \frac{-ln(30-5t-25 (e^{-1/5} - 1))}{ln(10)}$

$\frac{d(pH)}{dt} = -\frac{\frac{1}{30 - 5t - 25(e^{-1/5} - 1)} \cdot (-5)}{ln(10)} = \frac{5}{ln(10) \cdot (30 - 5t - 25(e^{-1/5} - 1))}$

Now plug in t = 10.
$\frac{d(pH)}{dt}(t = 10) = \frac{5}{ln(10) \cdot (30 - 50 - 25(e^{-1/5} - 1))} = \frac{5}{ln(10) \cdot (-20 - 25(e^{-1/5} - 1))}$

$= -\frac{1}{ln(10) \cdot (4 + 5(e^{-1/5} - 1))} \approx 0.140382$

-Dan
• July 26th 2007, 07:11 AM
curvature
Quote:

Originally Posted by topsquark
There's a solution missing from the above solved equation:
$0 =2x \ln x \left(1 + \ln x \right)$
Thus
$2x = 0$ ==> $x = 0$
or
$ln(x) = 0$ ==> $x = 1$
or
$1 + ln(x) = 0$ ==> $x = \frac{1}{e}$

However x=0 is not in the domain of the function. So x=1/e and x=1 are the answers.