# Equation of Tangent Line

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• Mar 27th 2013, 08:35 AM
hellokitty999
Equation 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 AM
joeDIT
Re: Equation of Tangent Line
is that meant to read $y= 5e^x (x^2+x-1)^{-x}$ in any case get what $y$ would be by substituting in $x=1$ then get $dy/dx$ of the original $y= 5e^x (x^2+x-1)^(-x)$ and then use the equation of a line formula [TEX]y-y_1=m(x-x_1) where $m=dy/dx$. Hope this helps.
• Mar 27th 2013, 10:29 AM
HallsofIvy
Re: 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.