Expert: Scotto - 5/15/2009

Question
QUESTION: How do I find the 3rd derivative of the second derivative that I got,
f''(x)= (-1/4)(t^2+1)^(-3/2)(4t^2)+(t^2+1)^(-1/2)

I've tried doing the product law with (-1/4)(t^2+1)^(-3/2)(4t^2) and adding (t^2+1)^(-1/2) to it, but i got it wrong. What do I do here??



This problem also,
I can't figure out the second derivative.. of the first,
I got f'(x)= 2t(e^(5t))+(5t^2)(e^(5t))


ANSWER: I need to see what the original function was.

Basically, if you have f²(g(h(x))), the deritvae is
2f(g(h(x))f'(g(h(x))g'(h(x))h'(x).

To send me the function, you could attach an image or
send it as set of other functions.  An example of how
the function could be put is
a(x) = 1/(t²+1)^1.5, b(x) = 4t², and c(t) = 1/(t²+1)^0.5.
The function might be f(x) = a(x)b(x) + c(x).

The function looks like it can be divided into f(x)=g(x)h(x).
Determine g'(x) and h'(x).  Then you can say that
f'(x) = g(x)h'(x) + h(x)g'(x).

Note that if one of the functions is of the form
g(x) = (m(x))^p, then g'(x) = (p*(m(x))^(p-1))*m'(x).

I await the full problem.


On the second one, the function appears to be t²e^(5t).
Look at each portion to get f"(t) from f'(t).
Let f(t) = a(t)b(t) + c(t)b(t).
The derivative is a'(t)b(t) + b'(t)a(t) + c'(t)b(t) + b'(t)c(t).

In this case, a(t) = 2t, b(t) = e^(5t), and c(t) = 5t².
From this, a'(t) = 2, b'(t) = 5e^(5t), and c'(t) = 10t.
After this is done, some terms can be combined.


---------- FOLLOW-UP ----------

QUESTION: I haven't finished reading your answer yet but the original functions was...

1. f(t)=sqrt(t^2+1)

2. f(t)=t^2(e^(5t))

For part 1, I need to find the 3rd deriv. f'''(t).
For part 2, I need to find the 2nd deriv f''(t).

For both parts I got up until the last deriv i needed to find.

2.

Answer
1. Give f(t) = (t²+1)^0.5, f'(t)=(0.5/(t²+1)^0.5)2t = t/√(t²+1).
Given this quotient rule, it can be seen that g(t) = t and
h(t) = (t²+1)^0.5.  From here, g'(t) = 1.

It can be seen that h(t) = f(t), so h'(t) = t/√(t²+1).

Given this, f"(t) = (hg' - gh')/h².

2. f(t) = g(t)h(t) where g(t) = t² and h(t) = e^(5t).
It is seen that g'(t) = 2t and h'(t) = 4e^(5t).
This means that f'(t) = gh' + g'h = 4t²e^(5t) + 2te^(5t).
We can factor out the e^(5t), so we have
f'(t) = e^(5t)(4t²+2t).  We could also factor out 2t,
but that would make taking the derivative a power of 3 functions.
I mean, this could be done, making f' = mno, but then f" would be
m'no + n'mo + o'mn.

Anyway, the last function we had was f'(t) = e^(5t)(4t²+2t).  
Here, g(t) = e^(5t), so g'(t) = 5e^(5t).
This time h(t) = 4t² + 2t, so h'(t) = 8t + 2.
This means that f"(x) = h'g + hg' = (8t+2)e^(5t) + (4t²+2t)5e^(5t).
Simplifying this gives us f"(t) = e^5t(8t + 2 + 4t²(5) + 2t(5)) =
e^5t(8t + 2 + 20t² + 10t) = e^5t(20t² + 18t + 2).