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derivative help!
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Im here.
\[e ^{1-2x}\]
Ok so the rule with e is that that g(x)=e^f(x) then g’(x)=f’(x)e^f(x)+C
f(x)=1-2x f’(x)=-2
so. . . -2e^1-2x +c?
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Yup.
ok. got it. you would't happen to know anything about natural logs would you? the problem is the derivative of \[\ln \sqrt{x-1}\]
g(x)=ln.f(x) g’(x)=f’(x)/f(x)
i got this. . . .
\[\frac{ \frac{ 1 }{ 2\sqrt{x-1} }}{ \sqrt{x-1} }\] which doesnt seem right
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Because f(x)=(x-1)^-1/2 f’(x)=-1/2(x-1)^-3/2
right. . .
so the answer is -1/2(x-1) how do we get rid of the ^-3/2?
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