Step 2 of 3We must now find the slope of the tangent line at the point of tangency, (1, In 2), by calculating the derivativeof s(t) and evaluating the derivative at t = 1.Recall the derivative of lIn x.1dIn x =dxUse the chain rule to find s'(t).s(t) = In(8 – 6t)s'(t)dt8 - 6tdtEvaluate s'(1).s'(1)SubmitSkip (you cannot come back) Tutorial ExerciseFind an equation of the tangent line at the point indicate.s(t) = In(8 6t),t = 1.Step 1 of 3Recall 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 ff(a) =(af '(a)The equation of the tangent line in point-slope form is y - f(a)f '(a) (x - a).To find the equation of the tangent line to the graph of s(t) = In(8 6t), we will need to find the slope of thetangent line to the graph of s(t) at the point (1, s(1)).Find s(1).s(1) = In(8 – 6(1)) = |In(2)In(2)(1, In(2)1, In(2)Therefore, the point of tangency is (t, s(t)) = (1, s(1)) =

Name: Step 2 of 3We must now find the slope of the tangent line at the poin
Uploaded: 2024-03-20T06:46:42.000Z
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Description: Step 2 of 3We must now find the slope of the tangent line at the point of tangency, (1, In 2), by calculating the derivativeof s(t) and evaluating the derivative at t = 1.Recall the derivative of lIn x.1dIn x =dxUse the chain rule to find s'(t).s(t)