Find the equation of the tangent line to the graph of y= 5e^x(x^2+x-1)^(-x) at the point where x=1

Printable View

- Mar 27th 2013, 08:35 AMhellokitty999Equation of Tangent Line
Find the equation of the tangent line to the graph of y= 5e^x(x^2+x-1)^(-x) at the point where x=1

- Mar 27th 2013, 10:00 AMjoeDITRe: Equation of Tangent Line
is that meant to read in any case get what would be by substituting in then get of the original and then use the equation of a line formula [TEX]y-y_1=m(x-x_1) where . Hope this helps.

- Mar 27th 2013, 10:29 AMHallsofIvyRe: Equation of Tangent Line
I presume you can do the derivative of 5e^x. To differentiate u= (x^2+ x- 1)^{-x} take the logarithm, ln(u)= -x ln(x^2+ x- 1) and differentiate both sides with respect to x.