Find an equation of the tangent line to the graph of y = e−x2 at the point (5, 1/e^25)

Question

Find an equation of the tangent line to the graph of y = e−x2 at the point (5, 1/e^25)

Answer 1

First find slope of tangent at point (5, 1/e^25) by differentiating given function.y = e^(-x^2)

dy/dx = (-2x) e^(-x^2) Evaluate dy/dx at x=5 gives slope m of tangent = -10 e^(-25)

In slope intercept form equation of tangent is y = mx +b , here m is slope and b is y-intercept

The equation is y = -10 e^(-25) x + b find intercept b , substitute point (5, 1/e^25) in the equation

1/e^(25) = -10 * 5/e^(25) + b solve for b

b = 51/e^(25)

Therefore equation of tangent is y = -10x/e^(25) + 51/e^(25)

I hope this helps. Please provide your comments.

Shailesh Kadakia, Senior Math and Physics Teacher

www.Skylativity.com