Question

Find the equation of the tangent line to the curve f(x)=ln(x^2) at the point x=e

Answer 1

first off, note that \(\ln(x^2)=2\ln(x)\) the take the derivative

Answer 2

what is the derivative of ln? e

Answer 3

you don't want the derivative of \(ln(e)\) which is just 1, you want the derivative of \(2\ln(x)\)

Answer 4

2

Answer 5

ok lets go slow

Answer 6

what is the derivative of \(\ln(x)\) ?

Answer 7

1/x

Answer 8

ok good

Answer 9

so the derivative of \(2\ln(x)\) is \(\frac{2}{x}\)

Answer 10

yes

Answer 11

now to find the slope at \(x=e\) replace \(x\) by \(e\) and get a slope of \(\frac{2}{e}\)

Answer 12

the point is \((x,2)\) and the slope is \(\frac{2}{e}\) find the equation of the line by using the point - slope formula

Answer 13

y-2=2/e x

Answer 14

where do u get the point (x,2)?