Question

Derivatives with square root

Answer 1

Lets do it one by one, okay?

Answer 2

Check out 7-10
I am having trouble figuring this out. My trouble is what do I do when i have x < 1 or something unknown like that. I also am having trouble with the derivative rule for sqrts

Answer 3

Oh, ok so the only thing you need help with is #7 through #10 ?

Answer 4

yup yup

Answer 5

\(\LARGE\color{black}{ f(x) = \begin{cases} & x+1,{~~~}{\large x<1} \\ & 2\sqrt{x},{~~~}{\large x\ge1} \end{cases} }\)

for the function to be differentiable at x=1, it first has to be continous at that point:
That means that: (1)+1=2√(1) ---> 2=2
So it is continous at x=1

Now, for the function to be differentiable, the instanaeous slope of one peace at x=1 and the instanenous slope of the other peace at x=1 must be equivalent. (we will do derivatives now)

Answer 6

What is the derivative of:

\([1]\) x+1
\([2]\) 2√x

Answer 7

What is that rule you just used to get (1) + 1= 2sqrt1

Answer 8

At that step I was just checking that the function is continous at x=1. For that to be true, both peaces have to be equivalent to each other when evaluated at x=1.

Answer 9

derivative of x+1 is 1

I am not sure how to do the one with sqrts

Answer 10

OHHHH
So its as simple as plugging in X

Answer 11

Ok,

\(\large\color{black}{ \displaystyle \frac{ d}{dx}2\sqrt{x} =\frac{ d}{dx}2(x)^{1/2} }\)

Answer 12

For continuity, yes it is as simple as plugging in x=1 to verify continuity at x=1.

But, for differentiablity we need to {take the derivative/find the slope-expression} of each peace and see whether the peaces have equivalent slopes at x=1 by plugging x=1 into the derivative of each peace.

Answer 13

OK so for 2sqrtx we get 2(x)^1/2. Is that the derivative?

Answer 14

No, I just wrote the square root differently.

Answer 15

2√x is same as 2(x)\(^{1/2}\)

Answer 16

Ok

Answer 17

Then?

Answer 18

Take the derivative:

\(\large\color{black}{ \displaystyle \frac{ d}{dx}x^n=nx^{n-1};{~~~~}n\ne1 }\)

Answer 19

Where did nx^n come from?

Answer 20

that is the rule you need, and when you are multiplying times constant c, is:

\(\large\color{black}{ \displaystyle \frac{ d}{dx}cx^n=cnx^{n-1};{~~~~}n\ne1 }\)

Answer 21

I am just trying to show some general rules of differentiation.

Answer 22

but it seems as though what I said hasn't made much sense, ain't it so?

Answer 23

I have a list of the rules but I am not sure how to apply them. This is what I am using http://www.mathsisfun.com/calculus/derivatives-rules.html#examples

Answer 24

So at this point, we know that it is continuous at 1

Answer 25

We want to know the derivative of 2sqrtx

Answer 26

yes, as soon as we evaluated both peaces at x=1 and saw that then they were equivalent to each other.

Answer 27

yes, we do...

Answer 28

The rule you should use is:

\(\large{\bbox[5pt, lightyellow ,border:2px solid black ]{ \displaystyle\frac{d}{dx}x^n=nx^{n-1} }}\)

Answer 29

\(\large{\bbox[5pt, lightyellow ,border:2px solid black ]{ \displaystyle\frac{d}{dx}x^{{\huge ½}}=\frac{1}{2}x^{{\huge ½}-1} }}\)

Answer 30

\(\large{\bbox[5pt, lightyellow ,border:2px solid black ]{ \displaystyle\frac{d}{dx}2\times x^{{\huge ½}}=2\times\frac{1}{2}x^{{\huge ½}-1} }}\)

Answer 31

So right now we have [2(x)^1/2]

Answer 32

yes, that is what you shall take the derivative of,.

Answer 33

same way as I already showed above.

Answer 34

Ok which rule is that last one called?

Answer 35

The power rule

Answer 36

All of this is the power rule

Answer 37

Is 2 our final answer? I tried to work it out