what is the 20th derivative of y=sin(2x)?
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
so f(x)=sin(2x) and f'(x)=?
2cos(2x)
\[f(x)=\sin(2x) \\ f^{(1)}(x)=2\cos(2x) \\ f^{(2)}(x)=-4\sin(2x)\] what is the 3rd derivative?
would it be -8cos(2x)?
\[f(x)=\sin(2x)=2^0\sin(2x) \\ f^{(1)}(x)=2\cos(2x)=2^1 \cos(2x) \\ f^{(2)}(x)=-4\sin(2x) =- 2^2 \sin(2x)\\ f^{(3)}(x)=-8 \cos(2x)=-2^3 \cos(2x) \]
so you should a couple of things in pattern
see*
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?
the degree of 2 is increasing
so we know that the power of 2 is going to be what for the 20th derivative?
for the first derivative we had 2^1 second we had 2^2 third we had 2^3 so 20th we had 2^?
20?
right
ohhhhh i seee
and lets determine if we need cos or sin
look at the even derivatives and the odd derivatives
20 is even what do you notice for all the even derivatives so far
they have been sin right?
so the 20th derivative would be correct to assume we will have sin
\[f^{(20)}(x)= (\text{ is this plus or minus) }2^{20}\sin(2x)\]
we almost are done with this
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
so 2^(20)*sin(2x)
This makes so much sense ! THANK YOU
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
but anyways I was just showing you one more for fun
did you want to see if you could find the 46th derivative?
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)
and no problem :)
Join our real-time social learning platform and learn together with your friends!