Find h′(2) given that f(2) = −1, f′(2) = 4, g(2) = 1, and g′(2) = −5. solve each one c. h(x) = f(x)/g(x). d. h(x) = ln f(x).
use the quotient rule for (c) and the chain rule for (d)
is the anserw to c 19
how did you get this answer?
this site might help you understand these rules better: http://www.1728.org/chainrul.htm
or would the anserw be 24 for c
it is best for you to actually show your work here rather than just giving out any final answers. that way I can help spot where you may have made a mistake.
\[h'(2)=f'(2)*g(2)-f(2)*g'(2)/g(2)^2\]
that looks correct - so what was your next step?
\[h'(2)=(4)(1)-(4)(-5)/(1)^2\] \[=4+20/1 \] so 4 +20 =24 and 24/1 =24
the second (4) is not correct - it should be the value of f(2) NOT f'(2)
ok so it should be 4-5/1 which will give you -1 Right?
that is correct
thsnks
thanks lol
yw :) hopefully you know how to use the chain rule for the second part?
well ln is 1/x and f(x) is 2 so would I do 1/2 since h'(2) and then times that by f'(x) which is 4 . not sure if its right.
take a look at the web site I linked to above, it also explains the chain rule further down the page. If you still don't understand it after reading that then let me know and I'll try to help.
I think the ln is whats confusing me for that problem
ok, let me do this one for you to illustrate how it works. we are given this:\[h(x)=\ln(f(x))\]lets substitute:\[y=f(x)\]so that:\[y'=f'(x)\]we then get:\[h(x)=\ln(y)\]therefore:\[\frac{dh}{dx}=\frac{dh}{dy}\times\frac{dy}{dx}=\frac{1}{y}\times f'(x)=\frac{f'(x)}{f(x)}\]
I hope that makes sense?
so I would take f'(x) and divided by f(x) ?
yes, h'(x) = dh/dx = f'(x)/f(x)
so the anserw would be -1 ?
please show your working...
\[f'(2)=4/f(2)=-1\] so 4/-1 = -4 my bad
correct :)
Thanks !
yw :)
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