Thread: Finding Derivatives of Functions

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!

Where is the problem? Do you know the derivative of ln(u) where u is a function? ...
You have to use the 'chain rule'.

I think that u= t^7-2 so therefore the derivative would be 7t^6

Originally Posted by StarDancer19

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.

What about t^8? ...
Why not: u=t^8(t^7-2)=t^15-2t^8?
You know:
D(ln(u))=D(u)/u

Can you finish it? ...

Yeah, I guess that the t^8 would also be apart of the u. But no, I'm still confused on how to finish it.

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).