# If f(x) = \cos(2x),\ then\ f'''(\pi/4)=?

If ?
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Step 1
Given function is $f\left(x\right)=\mathrm{cos}\left(2x\right)$.
Step 2
Compute the derivatives.
${f}^{\prime }\left(x\right)={\left(\mathrm{cos}\left(2x\right)\right)}^{\prime }=-2\mathrm{sin}\left(2x\right)$
$f{}^{″}\left(x\right)={\left(-2\mathrm{sin}\left(2x\right)\right)}^{\prime }=-2\left(2\mathrm{cos}\left(2x\right)\right)=-4\mathrm{cos}\left(2x\right)$
$f{}^{‴}\left(x\right)={\left(-4\mathrm{cos}\left(2x\right)\right)}^{\prime }=-4\left(-2\mathrm{sin}\left(2x\right)\right)=8\mathrm{sin}\left(2x\right)$
Obtain $f{}^{‴}\left(\frac{\pi }{4}\right)$.
$f{}^{‴}\left(\frac{\pi }{4}\right)=8\mathrm{sin}\left(2\left(\frac{\pi }{4}\right)\right)$
$=8\mathrm{sin}\left(\frac{\pi }{2}\right)$
=8(1)
=8