# How do you find the derivative of sin (cos(2t-5))?

May 26, 2016

$\frac{\mathrm{df}}{\mathrm{dt}} = - 2 \sin \left(2 t - 5\right) \cos \left(\cos \left(2 t - 5\right)\right)$

#### Explanation:

To find the derivative of sin(cos(2t−5)), we use the concept of function of a function and use chain rule to find derivative.

Here $f \left(t\right) = \sin t$, $g \left(t\right) = \cos t$ and $h \left(t\right) = \left(2 t - 5\right)$

Hence f(g(h(t)))=sin(cos(2t−5))

as $\frac{\mathrm{df}}{\mathrm{dt}} = \frac{\mathrm{df}}{\mathrm{dg}} \times \frac{\mathrm{dg}}{\mathrm{dh}} \times \frac{\mathrm{dh}}{\mathrm{dt}}$

Hence $\frac{\mathrm{df}}{\mathrm{dt}} = \cos \left(\cos \left(2 t - 5\right)\right) \times \left(- \sin \left(2 t - 5\right)\right) \times 2$

or $\frac{\mathrm{df}}{\mathrm{dt}} = - 2 \sin \left(2 t - 5\right) \cos \left(\cos \left(2 t - 5\right)\right)$