Mathematics OpenStudy (anonymous):

find the equation of the tangent line to the graph of y=f(x) at x=xknot . #) f(x)=ln x; x=e^-1 answer is "y=ex-2", but how do you do that? I need to know the steps. OpenStudy (anonymous):

yo ninjanick, were all here to help, but your question honestly isn't worded correctly at all. we don't know what you are asking. OpenStudy (anonymous):

find the tangent line to the graph OpenStudy (anonymous):

you didn't give us a graph or equation. is the equation ln(x) or e^-1. and at x=xknot doesn't mean anything either. OpenStudy (anonymous):

here is the function f(x)=ln(x) and x=e^-1, answer is suppose to be "y=ex-2", so they want us to find the equation of the tangent line of that function. OpenStudy (anonymous):

if the function is f(x) = ln(x) and x = e^-1, then it is not a function, it is simply ln(1/e), which is a number, not a function. Do you want the equation of the tangent line of f(x) = ln(x) at the point x=1/e? OpenStudy (dumbcow):

$f(x) = \ln x$ $f'(x) = \frac{1}{x}$ $x = e^{-1} \rightarrow f(e^{-1})=\ln(e^{-1}) = -1$ $m = f'(e^{-1}) = \frac{1}{e^{-1}} = e$ $y = mx + b$ $\rightarrow -1 = e*(1/e) + b$ $b = -2$ $\rightarrow y = e*x - 2$ OpenStudy (anonymous):

If that is the case, take derivative of ln(x) = i / x. At 1/e, we get the derivative = e. OpenStudy (anonymous):

yep there u go hope thats what ur asking for OpenStudy (anonymous):

thanks dumbcow, but you should be named smartcow.

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