A curve is given by the equation $\displaystyle y^3 + 220 = (e^x +1)^2 $ find the slope of the tangent line at point (0, -6)
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Originally Posted by viet A curve is given by the equation $\displaystyle y^3 + 220 = (e^x +1)^2 $ find the slope of the tangent line at point (0, -6) $\displaystyle \frac{d}{dx} [y^3 + 220] = \frac{d}{dx}(e^x +1)^2 $ so: $\displaystyle 3y^2 \frac{dy}{dx} = 2(e^x+1)e^x $ so at the required point: $\displaystyle 108 \frac{dy}{dx} = 4 $ RonL
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