y=e^-9t at t=0

Since the derivative y ' = -9e^(-9t) equals -9 at t = 0, we can write the equation of the tangent as y - 1 = -9t. The equation of the tangent to the graph y = y(t) at t = t_0 is y - y(t_0) = y ' (t_0) (t - t_0).Answer: y = 1 - 9t.

Since the derivative y ' = -9e^(-9t) equals -9 at t = 0, we can write the equation of the tangent as

y - 1 = -9t. The equation of the tangent to the graph y = y(t) at t = t_0 is y - y(t_0) = y ' (t_0) (t - t_0).

Answer: y = 1 - 9t.

how would i find the equation tot he tangent line y=6^x x=1

y(1) = 6; y ' (x) = 6^x * ln6; y '(1) = 6 ln6.

The equation is y - 6 = 6 ln6 (x - 1).

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