Confused on these two derivative problems

[Niaboc67]

Homework Statement

a. k(t) = (sqrt(t+1))/(2t+1)b. y = (3^(x^2+1))(ln(2))

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

 

[Qwertywerty]

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.

Take the LCM - it won't be simplified too much - but that's the way it's going to stay .

I don't understand where the e or ln(3) comes from. Or why it's put as an exponent.

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 ?

Also, why does the ln(2) stay stationary?

You're only raising 3^x2+1 to e .

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.

Do you know the chain rule ?

Hope this helps .

 

[SteamKing]

Homework Statement

a. k(t) = (sqrt(t+1))/(2t+1)b. y = (3^(x^2+1))(ln(2))

The Attempt at a Solution

For the first problem, I know I use the quotient rule for derivatives (L)(DH)-(H)(DL)/((L)^2)

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

 

[SteamKing]

Homework Statement

a. k(t) = (sqrt(t+1))/(2t+1)b. y = (3^(x^2+1))(ln(2))

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.

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 = x^1/2 ".

 

[HallsofIvy]

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

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 $f'(x)a^{f(x)} ln(a)$.