MHB 205.8.9 Find the derivative of the function

[karush]

205.8.9 Find the derivative of the function
$y=\cos(\tan(5t-4))\\$
chain rule $u=\tan(5t-4)$
$\frac{d}{du}\cos{(u)} \frac{d}{dt}\tan{\left(5t-4\right)}\\$
then
$-\sin{\left (u \right )}\cdot 5 \sec^{2}{\left (5 t - 4 \right )}\\$
replacing u with $\tan(5t-4)$
$-\sin{(\tan(5t-4))}\cdot 5 \sec^{2}{(5t-4)}\\$
$W\vert A$ returns

$-5\sec^2(5t-4)\sin(\tan(5t-4))$

ok I got confused on this u substitution thing but still seems to match the $W\vert A$ return

 

[MarkFL]

I would write:

$$y'=-\sin\left(\tan(5t-4)\right)\sec^2(5t-4)(5)=-5\sec^2(5t-4)\sin\left(\tan(5t-4)\right)\quad\checkmark$$

 

[karush]

Cool

I formerly ran this through eHm
but like 27 steps...

I really question these "show steps"
on online calculators ..
Forums much better

 

[HOI]

205.8.9 Find the derivative of the function
$y=\cos(\tan(5t-4))\\$
chain rule $u=\tan(5t-4)$
$\frac{d}{du}\cos{(u)} \frac{d}{dt}\tan{\left(5t-4\right)}\\$
then
$-\sin{\left (u \right )}\cdot 5 \sec^{2}{\left (5 t - 4 \right )}\\$
replacing u with $\tan(5t-4)$
$-\sin{(\tan(5t-4))}\cdot 5 \sec^{2}{(5t-4)}\\$
$W\vert A$ returns

$-5\sec^2(5t-4)\sin(\tan(5t-4))$

ok I got confused on this u substitution thing but still seems to match the $W\vert A$ return

It is a good idea to explictely write u= tan(5t-4) so that y= cos(u). I would go a step further and write v= 5t- 4 so that u= tan(v). Then the chain rule says that dy/dx= (dy/du)(du/dx)= (dy/du)(du/dv)(dv/dx)= (-sin(u))(sec^2(v))(5)= (-sin(tan(5t-4)))(sec^2(5t-4))(5).