# Thread: find slope of tangent line

1. ## find slope of tangent line

A curve is given by the equation $y^3 + 220 = (e^x +1)^2$
find the slope of the tangent line at point (0, -6)

2. Originally Posted by viet
A curve is given by the equation $y^3 + 220 = (e^x +1)^2$
find the slope of the tangent line at point (0, -6)
$\frac{d}{dx} [y^3 + 220] = \frac{d}{dx}(e^x +1)^2$

so:

$
3y^2 \frac{dy}{dx} = 2(e^x+1)e^x
$

so at the required point:

$
108 \frac{dy}{dx} = 4
$

RonL