I am working on some homework and am stuck when it comes to problems like these:
Find ds/dt if s=ln[t^8(t^7-2)]
Thanks for the help!
One way is to use $\displaystyle ln(ab)=lna+lnb$
$\displaystyle \frac{d}{dt}ln\left[t^8\left(t^7-2\right)\right]=\frac{d}{dt}lnt^8+\frac{d}{dt}ln\left(t^7-2\right)$
$\displaystyle x=t^8,\;\;y=t^7-2$
$\displaystyle \frac{ds}{dt}=\frac{d}{dt}lnx+\frac{d}{dt}lny$
$\displaystyle =\frac{dx}{dt}\;\frac{d}{dx}lnx+\frac{dy}{dt}\; \frac{d}{dy}lny$
You could also first evaluate $\displaystyle t^8\left(t^7-2\right)=t^{15}-2t^8$
and work from there.
You can follow what Archie meade said or:
Let u=t^8(t^7-2)=t^15-2t^8
So:
D(ln(u))=D(u)/u=(D(t^15-2t^8))(/t^15-2t^8)
If you can calculate this derivative: D(t^15-2t^8) then you are finished (you can also simplify the fraction).