- #1
karush
Gold Member
MHB
- 3,269
- 5
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
$y=\cos(\tan(5t-4))\\$
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