Question

Derivative help

Answer 1

I know d/dx for it is 1/2 x ^(-1/2)

Answer 2

And 3x^2 for part B.

Answer 3

what do you need help with exactly?
pluggin in x=1 into those?

Answer 4

multiply down and dubtract one from the exponent. Then rewrite as a fraction.

Answer 5

I guess I'm blind. The step is y'(1)=1/2.

Does it mean y=1/2 (1) ^(-1/2)

Answer 6

no y'=1/2(1)^(-1/2)=1/2

Answer 7

But 1/2 ^(-1/2) is not 1/2

Answer 8

why are you calculating 1/2 ^(-1/2)?

Answer 9

I don't understand why we are even plugging x=1 in. We need the derivative for slope. That's easy. But I do not get there the solution manual has a final answer of y'(1)=1/2.

Answer 10

right the slope of the tangnet line at (1,1) can be found by pluggin in the point (x=1,y=1) into the derivative of y=x^(1/2)

Answer 11

I know this is stupid, but will you do it out for me?

Answer 12

\[y'=\frac{1}{2}x^{\frac{-1}{2}} \\ y'|_{(x=1,y=1)}=\frac{1}{2}(1)^{\frac{-1}{2}}=\frac{1}{2}(1)=\frac{1}{2}\]

Answer 13

1 to any power is 1

Answer 14

if the negative exponent then write without negative exponents
\[y'=\frac{1}{2 x^{\frac{1}{2}}}\]

Answer 15

Ok. I need to make a note of that. I was multiplying 1/2 *1 first. Need to remember order of operations.

Answer 16

Thank you.

Answer 17

i hate when i leave words out
i do that all the time on the internet for some reason

if the negative exponent is confusing you*

Answer 18

hey @AmTran_Bus based on the first sentence it sounds like it also wants you to use the graph to estimate the slope

Answer 19

did you also do this?

Answer 20

I would just use the algebraic definition of slope to do this

Answer 21

This is actually for self study...its not an assignment. I'm just reviewing derivatives and working problems.

Answer 22

oh ok

Answer 23

So I'm not really worried bout it. But thanks so much!!!!!!!