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Mathematics 29 Online
OpenStudy (anonymous):

what is the 20th derivative of y=sin(2x)?

OpenStudy (freckles):

for these type of problems you need to find the pattern in the first few derivatives to make a guess for the nth derivative or in this case the 20th derivative

OpenStudy (freckles):

so f(x)=sin(2x) and f'(x)=?

OpenStudy (anonymous):

2cos(2x)

OpenStudy (freckles):

\[f(x)=\sin(2x) \\ f^{(1)}(x)=2\cos(2x) \\ f^{(2)}(x)=-4\sin(2x)\] what is the 3rd derivative?

OpenStudy (anonymous):

would it be -8cos(2x)?

OpenStudy (freckles):

\[f(x)=\sin(2x)=2^0\sin(2x) \\ f^{(1)}(x)=2\cos(2x)=2^1 \cos(2x) \\ f^{(2)}(x)=-4\sin(2x) =- 2^2 \sin(2x)\\ f^{(3)}(x)=-8 \cos(2x)=-2^3 \cos(2x) \]

OpenStudy (freckles):

so you should a couple of things in pattern

OpenStudy (freckles):

see*

OpenStudy (freckles):

It looks like every two will be positive, right? and also you notice we have the sin and cos switching with every term. And what else do you notice?

OpenStudy (anonymous):

the degree of 2 is increasing

OpenStudy (freckles):

so we know that the power of 2 is going to be what for the 20th derivative?

OpenStudy (freckles):

for the first derivative we had 2^1 second we had 2^2 third we had 2^3 so 20th we had 2^?

OpenStudy (anonymous):

20?

OpenStudy (freckles):

right

OpenStudy (anonymous):

ohhhhh i seee

OpenStudy (freckles):

and lets determine if we need cos or sin

OpenStudy (freckles):

look at the even derivatives and the odd derivatives

OpenStudy (freckles):

20 is even what do you notice for all the even derivatives so far

OpenStudy (anonymous):

they have been sin right?

OpenStudy (freckles):

so the 20th derivative would be correct to assume we will have sin

OpenStudy (freckles):

\[f^{(20)}(x)= (\text{ is this plus or minus) }2^{20}\sin(2x)\]

OpenStudy (freckles):

we almost are done with this

OpenStudy (freckles):

zero derivative=pos 1st derivative=pos 2nd derivative=neg 3rd derivative=neg so 20 divided by 4=5 remainder 0 the remainder 0 tells us that we have pos since the pattern repeats itself

OpenStudy (freckles):

so 2^(20)*sin(2x)

OpenStudy (anonymous):

This makes so much sense ! THANK YOU

OpenStudy (freckles):

if we had to find the 22nd derivative this would be \[f^{(22)}(x)=-2^{22}\sin(2x)\] 22/4=5 remainder 2 and recall the 2nd derivative=neg so that is how we got the 22nd derivative was going to be negative

OpenStudy (freckles):

but anyways I was just showing you one more for fun

OpenStudy (freckles):

did you want to see if you could find the 46th derivative?

OpenStudy (freckles):

even derivative we have sin odd derivative we have cos - we had 2^(the number of derivative we were on) - follow the pattern zero derivative=pos first derivative=pos second derivative=neg third derivative=neg to find the sign (neg or pos sign that is)

OpenStudy (freckles):

and no problem :)

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