find first three derivatives of f[x] = 2 cosx sin2x

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confused_07

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Please let me know if I am on the right path:

Find the first three derivatives of the function f[x]= 2cosx sin2x

f'[x]= (2cosx)(2cos2x) - (2sinx)(sin2x)
f"[x]= [(2cosx)(-2sin2x) + (-2sinx)(2cos2x)] - [(2sinx)(2cos2x) + (2cosx)(sin2x)]

I just want to make sure I am ok so far before I attempt the third derivative. Thanks
 
confused_07 said:
Please let me know if I am on the right path:

Find the first three derivatives of the function f[x]= 2cosx sin2x

f'[x]= (2cosx)(2cos2x) - (2sinx)(sin2x)
f"[x]= [(2cosx)(-2sin2x) + (-2sinx)(2cos2x)] - [(2sinx)(2cos2x) + (2cosx)(sin2x)]

I just want to make sure I am ok so far before I attempt the third derivative. Thanks
Click to expand...

In f''(x) you made a mistake in the first term in the square brackets:

(2cosx)(-2sin2x) should be (2cosx)(-4sin2x)

Note that it is easier to differentiate as follows. Instead of f(x) you repeatedly differentiate Ln[f(x)]. Then you solve for the derivatives of f(x) as follows. If g(x) =Ln[f(x)] then the chain rule and product rules give:

g'(x) = f'(x)/f(x)

g''(x) = f''(x)/f(x) - [f'(x)/f(x)]^2 = f''(x)/f(x) - g'(x)^2

The derivatives of g are very easy to evaluate and you can use these to calculate the derivatives of f.
 
You could try something else.

f(x) = 2cos(x)sin(2x) = 4[cos(x)]^2sin(x) = 4(sin(x) - [sin(x)]^3)

f'[x]= (2cos(x))(2cos(2x)) - (2sin(x))(sin(2x)) = cos(x)cos(2x) + 2cos(3x)

I'm not really making a firm recommendatino, here, just suggesting that you not forget your trigonometry. It may make your life easier, at times. In this case, I think your second derivative will be a lot easier with the form I suggested. The first derivative, not so much.

Keep up on those earlier skills!

You managed the first derivative, but the second wandered off just a bit. you missed the extra 2 from the chain ruls on the '2x'-derivatives.

Again, at least give it a shot at simplification before proceeding to the next level.

Good work. You did well for a first attempt.
 
it may be easier in your first derivative to write 2cos x (2cos 2x) as
4 cos x cos 2x, so you don't lose the 2's, as you did in the second derivative.
when you have coefficients in front of these terms and you have multiplication, combine them. example 5cos x (4sin x) = 20 cos x sin x
or 3 cos x (-2 sinx) = -6 cosx sinx.

you have the correct idea, just combine first then retry your second derivative.

Hopefully this helps.
 
Forgot to write down the formula for g'''(x):

g'''(x) = f'''(x)/f(x) -g'(x)[(g'(x))^2 + g''(x)] -2g'(x)g''(x)

We have g(x) = Ln(2) + Ln[cos(x)] + Ln[sin(2x)] and thus:

g'(x) = -tan(x) + 2cot(2x)

g''(x) = -1/[cos(x)]^2 -4/[sin(2x)]^2

g'''(x) = -2sin(x)/[cos(x)]^3 -16cos(x)/[sin(2x)]^3
 
I went the long route so I could see if I understood trigonometric derivatives, but I am sure there is a mistake with this much math:

F"[x] = (2cosx)(-4sin2x) - (2sinx)(2cos2x) - (2sinx)(2cos2x) + (2cosx)(sin2x)

. . . ..= (2cosx)(-4sin2x) - (4sinx)(4cos2x) + (2cosx)(sin2x)

F"'[x] = (2cosx)(-8cos2x) - (2sinx)(-4sin2x) - (4sinx)(-8sin2x) + (4cosx)(4cos2x) + (2cosx)(2cos2x) - (2sinx)(sin2x)

. . . . .= (-16cosx cos2x) + (8sinx sin2x) + (32sinx sin2x) + (16cosx cos2x) + (4cosx cos2x) - (2sinx sin2x)

. . . . .= (4cosx cos2x) + (40sinx sin2x) - (2sinx sin2x)

Holy moly! :shock:
 
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