Question

Find f '(x) for f of x equals the square root of the quantity sine of 8 times x.

a. f prime of x equals the quotient of the cosine of 8 times x and 2 times the square root of the quantity sine of 8 times x
b. f prime of x equals the quotient of 4 times the cosine of 8 times x and the square root of the quantity sine of 8 times x
c. f prime of x equals the quotient of 4 and the square root of the quantity sine of 8 times x
d. f'(x) = sin(4x)

Answer 1

@123AB456C @tkhunny

Answer 2

\[\large\rm f(x)=\sqrt{\sin8x}\]

Answer 3

Hey there :)
This is a handy derivative that you'll want to put to memory,\[\large\rm \frac{d}{dx}\sqrt x=\frac{1}{2\sqrt x}\]Derivative of square root is `one over two square roots`.
You can do your 1/2 power if you prefer.
Square root shows up so often though, that it's worth remembering :)

Answer 4

The derivative for f(x)=\sqrt{\sin 8x}\] is \[4\csc(8x)12\cos(8x)\]

Answer 5

Hmm where did your square root go? :O
It vanished!

Let's apply our square root derivative,\[\large\rm f(x)=\sqrt{\sin8x}\]so then,\[\large\rm f'(x)=\frac{1}{2\sqrt{\sin8x}}\cdot(\sin8x)'\]And then of course we have chain rule on the outside.

Answer 6

Your solution doesn't look far off,
just curious about a few things.

Answer 7

d/dxf(g(x))=f′(g(x))g′(x) isn't this the way to find a derivative using the chain rule?

Answer 8

Yes.
Does something I said not seem to match up with that? :o

Answer 9

No I was just making sure I was doing it right.

Answer 10

So our derivative of the inner function, our g(x)=sin8x, gives us\[\large\rm f'(x)=\frac{1}{2\sqrt{\sin8x}}\cdot(8\cos8x)\]

Answer 11

So I'm not quite sure where your 12 is coming from, hmm

Answer 12

okay after apply the chain rule to the problem you gave me I got \[2\csc(8x)\frac{ 1 }{ 2} \cos(8x)\]

Answer 13

you still there @zepdrix

Answer 14

Can you please help? @ Nnesha @triciaal I think @zepdrix left.

Answer 15

where did you get 12 from?
how did you simplify 1/(squareroot Sin 8x) to get

Answer 16

Ah sorry bout that D:

Answer 17

review what zepdrix has

Answer 18

it wasn't supposed to be 12. I think when I typed it in here I forgot the /. Its supposed to be 1/2

Answer 19

Oh oh I see, you intended for that 1/2 to be the exponent.
Ok this is making more sense :)

Answer 20

ok and he is back

Answer 21

ohh no problem @zepdrix. I have been having computer problems and I didn't know if is was from my end.

Answer 22

Keep in mind that the 2 should stay in the denominator,\[\large\rm \frac{1}{2(\sin8x)^{1/2}}\quad=\quad \frac{(\csc8x)^{1/2}}{2}\]

Answer 23

Yah I am new to Open study so I am still trying to figure out how to work everthing

Answer 24

now do I find the derivative of (csc8x) 1/2 2

Answer 25

Again, I would strongly recommend you try using the square root derivative :))
But if you like your 1/2 powers,
\[\large\rm f(x)=(\sin8x)^{1/2}\]just proceed carefully,\[\large\rm f'(x)=\frac12(\sin8x)^{-1/2}\cdot(\sin8x)'\]\[\large\rm f'(x)=\frac12(\sin8x)^{-1/2}\cdot(8\cos8x)\]

Answer 26

Notice with your definition of chain rule,
d/dx f(g(x)) = f'(g(x))*g'(x)

See how the f'( ) contains the `original inside function`,
the inner function doesn't change when you're applying this differentiation process.
Chain rule tells you to instead make a copy of the inner function,
and multiply by its derivative.

Answer 27

Okay so the derivative of the question using the chain rule is.
\[4\csc(8x)^\frac{ 1 }{ 2 }\cos(8x)\]

Answer 28

I am not sure how to use the square root derivative. I was only taught to use the chain rule derivative

Answer 29

I'm not sure where this 4 is coming from...
So from this step,\[\large\rm f'(x)=\frac12(\sin8x)^{-1/2}\cdot(\sin8x)'\]Yes, you can apply your trig identity to the sine,\[\large\rm f'(x)=\frac12(\csc8x)^{1/2}\cdot(\sin8x)'\]So you have a 1/2 in front, right?
And we'll need to fix up our sine derivative at the end as well.
One small step in chain rule that you're missing :)

Answer 30

wait why does your 1st problem have a -1/2 and you 2nd problem have a +1/2

Answer 31

Applying one of our exponent rules gives us this,\[\large\rm (\sin8x)^{-1/2}\quad=\quad\frac{1}{(\sin8x)^{1/2}}\]And then trig identity from there.

This is the exponent rule I'm thinking of:\[\large\rm x^{-a}=\frac{1}{x^a}\]

Answer 32

Ohh okay so now I need to find the derivative of your problem with the +1/2 correct?

Answer 33

Ah sorry, I mean, yes :)
you need to find the derivative of the thing that has the ' on it.

Answer 34

they both do lol

Answer 35

From this step,\[\large\rm f'(x)=\frac12(\csc8x)^{1/2}\cdot(\sin8x)'\]we need to find derivative of sin8x, that is our chain rule step.

Your cos8x was very close.

Answer 36

But it turns out that you have to chain again! :)\[\large\rm (\sin8x)'\quad=\quad (\cos8x)\cdot(8x)'\]

Answer 37

okay so apply the chain rule to \[(\cos 8x) \times (8x)\]

Answer 38

it's not just 8x on the outside,
it's the derivative of 8x,
don't forget the ' in that step! :)

Answer 39

I hope you understand this notation ok,
when I leave a tick mark on something,
it's just to "set up" the next step.

It's telling us that we still need to take a derivative.

Answer 40

\[\large\rm (\sin8x)'\quad=\quad (\cos8x)\cdot(8x)'\]Derivative of 8x? :)\[\large\rm \frac{d}{dx}8x=?\]

Answer 41

8

Answer 42

\[\large\rm (\sin8x)'\quad=\quad (\cos8x)\cdot(8)\]Good good good.

Answer 43

\[\large\rm f'(x)=\frac12(\csc8x)^{1/2}\cdot(\sin8x)'\]\[\large\rm f'(x)=\frac12(\csc8x)^{1/2}\cdot(\cos8x)\cdot(8x)'\]\[\large\rm f'(x)=\frac12(\csc8x)^{1/2}\cdot(\cos8x)\cdot(8)\]Bring the 8 and 1/2 together as a final step.
Multiply them.

Answer 44

Hmm, I didn't realize you had multiply choice options, woops.
Notice that `all of your options` involve sin8x, not csc8x.
So we probably should not have used that trig property.\[\large\rm f'(x)=\frac12(\sin8x)^{-1/2}\cdot(\cos8x)\cdot(8)\]

Answer 45

okay so do you want me to derivative that problem?

Answer 46

No derivative is all done :)
We did the hard part.

Now is just simplification.

Answer 47

Okay so what would be the steps? I am trying to follow put I am kindof having a hard time. Could you draw the steps on how to simplify

Answer 48

\[\large\rm f'(x)=\frac12(\sin8x)^{-1/2}\cdot(\cos8x)\cdot(8)\]Let's reapply that exponent step,\[\large\rm f'(x)=\frac12\frac{1}{(\sin8x)^{1/2}}\cdot(\cos8x)\cdot(8)\]Multiplication is `commutative`, so we can multiply in any order.
Let's bring the 8 to the front, and divide it by the 2,\[\large\rm f'(x)=\frac82\frac{1}{(\sin8x)^{1/2}}\cdot(\cos8x)\]Gives us,\[\large\rm f'(x)=4\frac{1}{(\sin8x)^{1/2}}\cdot(\cos8x)\]

Answer 49

And then as a final step, maybe put the stuff into the numerator,\[\large\rm f'(x)=\frac{4(\cos8x)}{(\sin8x)^{1/2}}\]

Answer 50

Oh maybe another good step would be to rewrite your 1/2 power as a square root again :)\[\large\rm f'(x)=\frac{4(\cos8x)}{\sqrt{\sin8x}}\]

Answer 51

oh okay now that makes sense. I was wondering what you were supposed to do with the 1/2 power. But now it makes sense.

Answer 52

Try to commit a lot of time to understanding/practicing the chain rule.
It is much more complex than the other derivative rules.
Blessings to you!

Answer 53

Thank you so very much for all the help. I got the answer CORRECT!!