If the tangent of the curve y = e ^ x is made through the origin, the tangent point coordinate is, and the tangent slope is Sorry, I didn't play well just now, hee hee
1. The derivative of y = e ^ x is y = e ^ X
2. So the tangent through (x0, Y0) is y = e ^ x0 (x-x0) + e ^ X
3. Because it passes through the origin (0,0), so 0 = e ^ x0 * (- x0) + e ^ x, x0 = 1
4. So the tangent is y = ex
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