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.

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.

find the tangent line to the graph

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.

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.

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?

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

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

yep there u go hope thats what ur asking for

thanks dumbcow, but you should be named smartcow.

Join our real-time social learning platform and learn together with your friends!