Question

Help with finding the tangent line. Equation in comments

Answer 1

well rewrite the function as

\[y = 3x^{-\frac{1}{3}}\]

then find the derivative

this will give the equation of the slope.

that's the 1st part...

hope it makes sense

Answer 2

Hey Katie c:
What part are you stuck on?
Do you understand how to find the derivative of this function?
As Camp suggested, rewriting it with a rational exponent might be easier to work with.
\[\Large\rm y=\frac{3}{\sqrt x}\qquad \to\qquad y=3x^{-1/2}\]
You'll be able to apply your power rule from there.

Answer 3

I don't understand any of it. None of the online lectures make sense to me.

Answer 4

aw :c

Answer 5

Hmm where to start then... thinkingggg... 0_0

Answer 6

So you remember your basic slope formula?\[\Large\rm m=\frac{y_2-y_1}{x_2-x_1}\]We write it a little different in big boy math,\[\Large\rm m=\frac{f(x+h)-f(x)}{h}\]This represents the slope of a `secant line`, a line through two points.
When we do this:\[\Large\rm \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\]We're letting the space between the points get smaller and smaller.
The limiting process let's the points get immeasurably close to one another to where you can think of it as a single point.

Now our slope value is that of a tangent line, a line touching the curve in one specific place.

Maybe this is a little too much info.. but anyway...

Answer 7

Our derivative shortcut rules allow us to skip this limiting process.
The derivatives always follow these nice patterns.

Answer 8

So, for example:
\[\Large\rm f(x)=x^2\]If we wanted to find the slope of the line tangent to the curve x^2 at x=3,
we would take it's derivative because the derivative function tells us about tangent slopes.

\[\Large\rm f'(x)=2x^1\]Power rule tells us to bring the power down in front,
and then decrease the power by 1.

So in this example,
the slope of our tangent line at x=3 would be:\[\Large\rm f'(3)=2(3)=6\]

Answer 9

What do you think miss katie? Too confusing still? D:

Answer 10

No it's actually making sense lol

Answer 11

The problem they gave you is a little trickier.
You have to remember some of your `root` and `exponent` stuff.

Answer 12

We want to use these two ideas:\[\Large\rm \frac{1}{x^1}=x^{-1}\]We can flip it up to the numerator by changing the sign of the exponent.

and:\[\Large\rm \sqrt[a]{x}=x^{1/a}\]

Answer 13

Recall that when no value is given to the root,
then it's a 2 by default.\[\Large\rm \sqrt{x}=\sqrt[2]{x}=x^{1/2}\]

Answer 14

Yeah I remember this stuff from pre calc

Answer 15

Ok good c:
So using those two rules,
\[\Large\rm y=\frac{3}{\sqrt x}=\frac{3}{x^{1/2}}=3x^{-1/2}\]We're able to write our function like that, yes?

Answer 16

It will be easier to deal with from here.
We can apply our power rule.
The -1/2 will come down in front as a multiplier.
And then we subtract 1 from the exponent.

Answer 17

\[\Large\rm y'=3\cdot\left(-\frac{1}{2}\right)x^{-\frac{1}{2}- 1}\]

Answer 18

Okay why are we multiplying by that?

Answer 19

That's what our power rule tells us to do.
Whatever is in the exponent, bring it down in front and multiply,
then subtract 1 from the exponent.\[\Large\rm \frac{d}{dx}x^n=n x^{n-1}\]

Answer 20

Okay makes sense

Answer 21

This is one of those things you need to get used to with negative exponents.
Like if you had a -2 exponent, you might expect it to go to -1 when you apply your power rule.

But your power rule tells you to subtract 1,
so when you have negative exponent, the NUMBER doesn't get smaller, it get's larger because it's becoming more negative.

Example:\[\Large\rm \frac{d}{dx}x^{-2}=-2x^{-3}\]

Answer 22

So after simplifying we get this:\[\Large\rm y'=-\frac{3}{2}x^{-3/2}\]

Answer 23

Okay I get it now thank you so much.

Answer 24

So for the first part of your question,
all they want you to do from there is plug in x=a.