Find derivative

[cainemvhzc]

Let f(x) = g(2sinx), where g′(√2) = √2. Find f′(π/4).

 

[Ishuda]

Let f(x) = g(2sinx), where g′(√2) = √2. Find f′(π/4).

Use the chain rule for f'(x). Then, what is x if
2 sin(x) = \(\displaystyle \sqrt{2}\)?

 

[Mathimatician]

2 many root 2's

I was confused at first at how to interpret... g'(√2)=√2 note: I have chosen to place one √2 in italics and the other √2 underlined.

But if you chain rule f(x) (as Ishuda Suggested) you get f'(x)=g'(2sinx)2cosx.

Now I thought to build the equation √2 = 2sinx because both expressions appear as inputs to g'.

√2 = 2sinx --> x = π/4, now let's remember that: 2sin(π/4) = √2 .

Bring f'(π/4) back into the discussion.

f'(π/4)=g'(2sin(π/4))2cos(π/4)... [now, because, 2sin(π/4) = √2 ]

f'(π/4)=g'(√2)2cos(π/4)... [and because, g'(√2)=√2]

f'(π/4)= √2(2cos(π/4))... [and because, cos(π/4) = √2/2]

f'(π/4)= √2(2(√2/2))

f'(π/4)= 2

Is this reasoning ok?

 

[Ishuda]

I was confused at first at how to interpret... g'(√2)=√2 note: I have chosen to place one √2 in italics and the other √2 underlined.

But if you chain rule f(x) (as Ishuda Suggested) you get f'(x)=g'(2sinx)cosx.

Now I thought to build the equation √2 = 2sinx because both expressions appear as inputs to g'.

√2 = 2sinx --> x = π/4, now let's remember that: 2sin(π/4) = √2 .

Bring f'(π/4) back into the discussion.

f'(π/4)=g'(2sin(π/4))cos(π/4)... [now, because, 2sin(π/4) = √2 ]

f'(π/4)=g'(√2)cos(π/4)... [and because, g'(√2)=√2]

f'(π/4)= √2cos(π/4)... [and because, cos(π/4) = √2/2]

f'(π/4)= √2√2/2

f'(π/4)= 1

Is this reasoning ok?

That's the way I would have gone about it except I would have
f'(x) = g'(2 sin(x)) (2 sin(x))' = g'(2sin(x)) 2 cos(x)
and come up with a slightly different answer

 

[Mathimatician]

oops I missed that 2,

thanks Ishuda, I went back and edited my original post.

 

[Steven G]

Use the chain rule for f'(x). Then, what is x if
2 sin(x) = \(\displaystyle \sqrt{2}\)?

I must be missing something or maybe the question has been changed. The op wants to know what y' equals for some x value. So why are we solving for x?
I would think that f '(x)= g'(2sinx)2cosx and f '(pi/4)= 2 and we are done.

 

[Ishuda]

I must be missing something or maybe the question has been changed. The op wants to know what y' equals for some x value. So why are we solving for x?
I would think that f '(x)= g'(2sinx)2cosx and f '(pi/4)= 2 and we are done.

You can either go forward
2 sin( x ) = 2 sin(\(\displaystyle \pi\) / 4) =2 / \(\displaystyle \sqrt{2}\) = \(\displaystyle \sqrt{2}\)

Or you can go backward, essentially asking what x satisfies
2 sin( x ) = \(\displaystyle \sqrt{2}\)
or
x = arcsin(\(\displaystyle \sqrt{2}\) / 2) = \(\displaystyle \pi\) / 4

Being somewhat a backwards sort of person, I went backward.

\(\displaystyle \underset{\text{slap! slap! no no no don't feed the trolls}}{.}\)

 

[Steven G]

You can either go forward
2 sin( x ) = 2 sin(\(\displaystyle \pi\) / 4) =2 / \(\displaystyle \sqrt{2}\) = \(\displaystyle \sqrt{2}\)

Or you can go backward, essentially asking what x satisfies
2 sin( x ) = \(\displaystyle \sqrt{2}\)
or
x = arcsin(\(\displaystyle \sqrt{2}\) / 2) = \(\displaystyle \pi\) / 4

Being somewhat a backwards sort of person, I went backward.

\(\displaystyle \underset{\text{slap! slap! no no no don't feed the trolls}}{.}\)

Thanks. I have enough trouble going forward so I would never think of going backward.