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Find the equation of the tangent line at the given value y equals ln x squared at x equals 1?
Wiki User
∙ 15y ago
Given y=LN(x2). To find the equation of a tangent line to a point on a graph, we must figure out what the slope is at that exact point. We must take the derivitave with respect to x to come up with a function that will give the slope of the tangent line at any point on the graph. Any text book will tell you that the derivative with respect to X of LN(U) is the quantity U prime over U(In other words, the derivitive of U over U. U is whatever is inside the natural log). In this case, it is y = (2x/x2). To find the slope, we then plug in an x coordinate, 1. We get y=2. The slope of the tangent line at this point is 2. We now have m for the slope-intercept form y = mx + b. Now we must find b. B is the y intercept, aka where does the graph intersect the y axis. We use the slope. We know that the graph of the tangent line contains the point (0,1). We go down 2 and left one, to find that the tangent line intercepts the y axis at -1. We now have B. The equation of the tangent line of y = LN(x2) at X = 1 is y = 2X-1. Hope this helps!!!
Wiki User
∙ 15y ago
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5
