# Confused on these two derivative problems

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1. Aug 6, 2015

### Niaboc67

1. The problem statement, all variables and given/known data

a. k(t) = (sqrt(t+1))/(2t+1)

b. y = (3^(x^2+1))(ln(2))

3. The attempt at a solution
For the first problem, I know I use the quotient rule for derivatives (L)(DH)-(H)(DL)/((L)^2)
which would go to: ((2t+1)(1/(2sqrt(t+1)) - (sqrt(t+1))(2))/((2t+1)^2) I get stuck here, maybe it's the algebra but I don't know how to factor all this down to get a solution.

For the second problem I was able to get the answer online but I don't understand the answer.
It goes:

y = (3^(x^2+1))(ln(2))
refine y = e^((ln(3))(x^2+1)) ln(2)
y' = [2xln(3)ln(2)]3^(x^2+1)

I don't understand where the e or ln(3) comes from. Or why it's put as an exponent. Also, why does the ln(2) stay stationary? Then e disappears as x^2+1 is broken into its derivative, the ln(3) comes down while 3^(x^2+1) is still put on the side. Could someone please explain this process works.

Thank you

2. Aug 6, 2015

### Qwertywerty

Take the LCM - it won't be simplified too much - but that's the way it's going to stay .
This is just the manipulation of the original question in order to convert into terms which you can solve by chain rule . Do you know basic logarithmic properties ?
You're only raising 3x2+1 to e .
Do you know the chain rule ?

Hope this helps .

3. Aug 6, 2015

### SteamKing

Staff Emeritus
A suggestion:

Rather than make up complicated expressions for derivatives, learn what notation is used normally and follow that:

y = f(x)

dy/dx = y' = f'(x) (use the ' to indicate differentiation w.r.t. x. The expression f'(x) is read, "f-prime of x".)

then, expressing the quotient rule in prime notation,

y = f(x) / g(x)

y' = (f(x)g'(x) - f'(x)g(x)) / [g(x)]2

4. Aug 6, 2015

### SteamKing

Staff Emeritus
When working problems involving taking derivatives, I find it easier to convert expressions like "sqrt (x)" or "√x " to exponential form, so that "sqrt(x) = √x = x1/2 ".

5. Aug 6, 2015

### HallsofIvy

You should know that $a^x= e^{ln(a^x)}= e^{x ln(a)}$. As result of that, the derivative of $a^x$ is $a^x ln(a)$ and the derivative of $a^{f(x)}$ is $f'(x)a^{f(x)} ln(a)$.