I must assume that you meant t=1 (not t=?1). If t=1, here's what we'd do:
1. Find the x and y values corresponding to t=1. They are:
x=(1)^7+1=2 and y=(1)^8+1=2. (Please note: write t^8 instead of t8, and write t^7 instead of t7.)
2. The slope of the tangent line to the graph is
dy/dt 8t^7+1
dy/dx = ---------- = ---------------- with 1 substituted for t
dx/dt 7t^6
Thus, dy/dx (at t=1) = 9/7
3. Now we have both a point (2,2) on the graph and the slope of the tangent line to the curve at that point: 9/7
4. The tangent line to the curve at (2,2) is found by using the point-slope formula:
y-y1 = m(x-x1)
which comes out to y-2 = 9/7(x-2), or 7y-14 = 9(x-2). You could, if you wished, simplify this result further (e. g., by solving for y in terms of x).
