1. The problem is to find the equation of the tangent line to the curve given by \(y = x^5 - x^3 + 2\) at the point where \(x = 1\). 2. First, find the derivative \(\frac{dy}{dx}\), which gives the slope of the tangent line at any point \(x\). $$\frac{dy}{dx} = 5x^4 - 3x^2$$ 3. Calculate the slope at \(x=1\): $$m = 5(1)^4 - 3(1)^2 = 5 - 3 = 2$$ 4. Find the y-coordinate of the point on the curve at \(x=1\): $$y = (1)^5 - (1)^3 + 2 = 1 - 1 + 2 = 2$$ 5. Use the point-slope form of a line equation \(y - y_1 = m(x - x_1)\) with \(m = 2\) and point \((1, 2)\): $$y - 2 = 2(x - 1)$$ 6. Simplify the equation: $$y = 2x - 2 + 2 = 2x$$ So, the equation of the tangent line is \(y = 2x\).