Take the function

$f(x) = \sin(2x) \cdot \cos(x)$

Find the 2013 th derivative

What I have found so far:

$f''(x)= -4\sin(x) \cdot \cos(2x) - 5f(x)$

I am assuming I need to find a relationship such as the above in order to just apply it again and again, however I can't seem to do so.

Any help is appreciated. Thank You.


EDIT: @Ihsan and Babak

Utilising this result, results in:

$f''(x) = -8\sin(3x) - f(x) $

And I'm not sure how to make recur this 1006 times!

  • 3
    $\begingroup$ Use the hints of Ihsan/Babak and the fact that $f(x)=g(x)+h(x)$ implies $f^{(n)}(x)=g^{(n)}(x)+h^{(n)}(x)$. $\endgroup$ – Hagen von Eitzen Feb 11 '13 at 10:35
  • $\begingroup$ Now note that $\frac {d^4}{dx^4} \sin (ax) = a^4 \sin (ax)$ $\endgroup$ – Ross Millikan Feb 11 '13 at 19:33

$$2\sin(2x)\cos(x)=\sin(3x)+\sin(x)$$ Use this.

  • $\begingroup$ edited my post. $\endgroup$ – Sy123 Feb 11 '13 at 10:30

Hint: Use $$\sin\alpha\cos β= \frac{1}2[\sin (\alpha + β) + \sin (\alpha − β)]$$ and $$\sin^{(n)}(x)=\frac{d^n}{dx^n}\sin(x)=\sin(x+\frac {\pi}{2}n)$$ and use the Chain rule also.

  • $\begingroup$ edited my post. $\endgroup$ – Sy123 Feb 11 '13 at 10:30
  • $\begingroup$ @Sy123: Do you want $f''(x)= -4\sin(x) \cdot \cos(2x) - 5f(x)$ becomes $f''(x) = -8\sin(3x) - f(x) $ above?? $\endgroup$ – Mikasa Feb 11 '13 at 10:35
  • $\begingroup$ Thank you, that really helped. (the nth derivative bit) $\endgroup$ – Sy123 Feb 11 '13 at 10:36
  • $\begingroup$ Your guidance is helpful! +1 $\endgroup$ – amWhy Feb 11 '13 at 17:36

Hint: Try to express the function using $e^x$ using: $\sin(\alpha) = \frac{e^{i \alpha} + e^{i\alpha}}{2i}$ and $\cos (\alpha) = \frac{e^{i\alpha}+e^{-i\alpha}}{2}$.

Full solution: (up to minor errors, if any...) $$ \sin(2x) \cos(x) =\left(\frac{e^{i2x}-e^{-i2x}}{2i} \right)\left(\frac{e^{ix}+e^{-ix}}{2} \right)=\frac{1}{4i}(e^{i3x}-e^{-i3x}+e^{ix}-e^{-ix}). $$ Since $(e^{ax})'=ae^{ax}$, the $2013$-th derivative is $$ (e^{ax})^{[2013]}=a^{2013}e^{ax}. $$ Plugging this to the former equation and recalling that $i^{2013}=i^{2012}i=1^{503}i=i$, we get $$ \begin{array}{lll} (\sin(2x) \cos(x))^{[2013]}&=&\left(\frac{1}{4i}(e^{i3x}-e^{-i3x}+e^{ix}-e^{-ix})\right)^{[2013]}\\ &=&\frac{1}{4i} \big((i3)^{2013} e^{i3x}- (-i3)^{2013}e^{-i3x}+(i)^{2013} e^{ix}- (-i)^{2013}e^{-ix}\big)\\ &=&\frac{i}{4i}\big( 3^{2013}(e^{i3x}+e^{-i3x}) + e^{ix}+e^{-ix}\big)\\ &=&\frac{3^{2013}}{2}\frac{e^{i3x}+e^{-i3x}}{2} +\frac{1}{2}\frac{e^{ix}+e^{-ix}}{2}\\\ &=&\frac{3^{2013}}{2}\cos 3x+\frac{\cos x}{2}. \end{array} $$


Note the 2nd derivative of $\sin(a x)$ is $-a^2 \sin (a x)$ (and similar for $\cos a x)$ that is $\frac{d^2}{dx^2} f(a x) = -a^2 f(a x)$ when $f(x)$ is sine or cosine. Doing two more derivatives you find the 4th derivative is $a^4 \sin( a x)$. Repeating this four more times, you get the 8th derivative is $a^8 \sin (a x)$ as eight is a multiple of 4. (Similarly 2012 is a multiple of 4; and then you have to do one more derivative to get to the 2013th derivative).

It helps to use trigonometric formulas to break the product of two trig functions into the sum of trig functions.


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