Question

Find the derivative of:
sqrt(3t+1)

Using h-> 0

Answer 1

I messed up somehow and got 3.

The answer is:

3
----------
2 sqrt(x+1)

Answer 2

are you supposed to use
\[ \lim_{h \rightarrow 0} \frac{ \sqrt{3(t+h)+1} -\sqrt{3t+1} }{h}\]

Answer 3

Yes.

thus far I got:

3
-----------------------
sqrt(3t+1) + sqrt(3x+1)

Answer 4

if so, the "trick" is to multiply top and bottom by the conjugate (i.e. same expression but with a plus sign in-between the terms)

Answer 5

Not sure if I made a mistake at this point.

Answer 6

that bottom looks wrong.
I would expect it to be
\[ \sqrt{3(t+h)+1} +\sqrt{3t+1} \]

Answer 7

you seem to be flipping between using "t" and "x" as the variable.

but after multiplying top and bottom by the conjugate, and simplifying you have
\[ \lim_{h \rightarrow 0} \frac{ 3}{ \sqrt{3(t+h)+1} +\sqrt{3t+1} } \]

Answer 8

I have 7 min to show my work then go to class

Answer 9

the last step is to let h go to zero.
that makes the bottom 2 * sqrt(stuff)

Answer 10

\[ \lim_{h \rightarrow 0} \frac{ 3}{ \sqrt{3(t+h)+1} +\sqrt{3t+1} } \\
\frac{ 3}{ \sqrt{3t+1} +\sqrt{3t+1} }
\]

Answer 11

thank you