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

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- 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 $\displaystyle y= 5e^x (x^2+x-1)^{-x}$ in any case get what $\displaystyle y$ would be by substituting in $\displaystyle x=1$ then get $\displaystyle dy/dx$ of the original $\displaystyle y= 5e^x (x^2+x-1)^(-x)$ and then use the equation of a line formula [TEX]y-y_1=m(x-x_1) where $\displaystyle m=dy/dx$. 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.