<p>I was also stumped by this one:</p>
<p>p(t)=ln |5t^5 + e^t|
find p'(t)</p>
<p>The book answer is (25t^4 + e^t)/(5t^5 + e^t), but I have no idea how they got it.</p>
<p>the general form of the derivative of ln(x) is 1/x dx, so...</p>
<p>= (1/5t^5+e^t) x (25t^4 + e^t), which is what the answer in your book is...</p>
<p>Thanks</p>
<p>I'm still a little confused, though. What about the absolute value?</p>
<p>-Rootbeer</p>
<p>You don't need the absolute value because there is no way to get a negative value from the derivative no matter what value of t you use.</p>
<p>Not quite, big al. You can still get negative numbers from the derivative (by way of the t^5 term).</p>
<p>The absolute value is there to control for the domain of the ln(x) function. While we can't take the ln(-5) we can take the ln|-5|. This absolute value allows us to not have to worry about domain too much in a problem like this.</p>
<p>There's also a more formal way to approacnh the problem, although WindSlicer is completely right. An expression with a square root can always be broken down into two separate expressions -- in this case, those are</p>
<p>p(t)=ln -(5t^5 + e^t) __<strong><em>5t^5 e^t <=0
and
p(t)=ln +(5t^5 + e^t) _</em></strong>_5t^5 e^t >=0</p>
<p><= and >= are meant to be less than or equal to, etc., by the way.</p>
<p>You can then take the derivatives of both expressions individually. You should get:</p>
<p>p'(t)= -(25t^4 + e^t) / -(5t^5 e^t) <strong><em>5t^5 e^t <=0
p'(t)= (25t^4 + e^t) / (5t^5 e^t) _</em></strong>5t^5 e^t >=0</p>
<p>If you look at those carefully, you should see that the negatives cancel out in the first expression. Thus, you can combine those two expressions, and get</p>
<p>p'(t)= (25t^4 + e^t) / (5t^5 e^t) t is a real number.</p>
<p>You don't have to go through that process each time you do this type of problem, but it might help you see why you can ignore the square root. In general, this method is useful to know because you may have to use it in other derivative problems involving absolute values. As you can see, the negative signs canceling out will not always happen -- natural logarithms just happen to work out nicely in that respect.</p>
<p>Does anyone have an official released exam by CB? If so, would you post the cutoffs for a 5,4? Someone said that 66/108 was a 5 but that seems really really lenient... just making sure.</p>