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I need to find the equation of the tangent line at the point: s(t)=Int, at t=5
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the equation from a straight line from algebra is\[y-y_1=m(x-x_1)\]where m is the slope, and y is the value of the function at that point since we know the slop at a point is given by the derivative at that point, we can make a calculus version of this formula for the equation of the tangent at some point:\[y-f(x_1)=f'(x_1)(x-x_1)\]so what is \[f'(x)\] in your case?
1/t^In
? no that's not\[\frac d{dt}\ln t\]try again please
Int/sqtr t
The derivative of ln(t) is 1/t....which gives you the slope of a tangent line at a point on the curve ln(t). It's then just a matter of algebra to find the equation for the tangent line given a point.
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