Find an equation of the tangent line at the point indicated.
s(t) = ln($a - $b t), t = 1
Recall that a tangent line to a graph of y = f(x) at a point P(a, f(a)) is the line through the point P of slope A283TCLjCmnifO4wSMIVtL2Tw0eo9gufYL5fTQ==. The equation of the tangent line in point-slope form is y - gtOou3Pq6m9e+Qz4TZ4fDaJmEUf+6R7MlMU6tg== = A283TCLjCmnifO4wSMIVtL2Tw0eo9gufYL5fTQ==ยท(x - a).
To find the equation of the tangent line to the graph of s(t) = ln($a - $b t), we will need to find the slope of the tangent line to the graph of s(t) at the point (1,s(1)).
Find s(1).
s(1) = ln($a - $b(1)) = ln(QYX0X2yMbpA=)
Therefore, the point of tangency is (t,s(t)) = (1,s(1)) = (1,ln(QYX0X2yMbpA=).
We must now find the slope of the tangent line at the point of tangency, (1,ln($bma)), by calculating the derivative of s(t) and evaluating the derivative at t = 1.
Recall the derivative of ln(x).
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Use the chain rule to find s'(t).
a = iSba6t70dtA= and b = nc1ItEz0kR4=
Evaluate s'(1).
c = QYX0X2yMbpA=
Write the equation of the tangent line.
m = 0VV1JcqyBrI=, n = QYX0X2yMbpA=