Find an equation of the tangent line to the graph of the function at the given point.
I was given y= (1/3)arccosx and derived it to y=-1 / 3 (1-x^2)^1/2
I don't know what to do next. Please tell me the steps I need to take. I'm not giving the points because I don't just want an answer, I want to know how to do it. Thanks in advance.
I was given y= (1/3)arccosx and derived it to y=-1 / 3 (1-x^2)^1/2
I don't know what to do next. Please tell me the steps I need to take. I'm not giving the points because I don't just want an answer, I want to know how to do it. Thanks in advance.
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I forget the inverse trig derivatives, so I'm going to go ahead and do this assuming you derived correctly.
If you'll recall, the derivative is the slope of the tangent line at every point. Therefore, by plugging in the x value of the specified point into your derivative, you'll get the SLOPE of the tangent line at that point. We'll call this result m.
Now, here's the point-slope form of a line, given point (x1, x2) and slope m
y-y1 = m(x - x1).
To get the point (x1, y1), we'll use the given point's x and y values, since we know these will be on the graph of the tangent line. If you're only given an x-value, plug it into the ORIGINAL equation to get the y-value.
So now you have an m, which you got by plugging the x-coordinate of the point into the derivative, and a point (x1, y1), which is just the point they gave you. This is the slope of the tangent line and the point at which the line is tangent to the graph, which is only significant because it guarantees this point will be on the tangent line.
y - y1 = m (x - x1)
Plug your values into the point-slope form of a line and simplify to get the equation of the tangent line.
If you'll recall, the derivative is the slope of the tangent line at every point. Therefore, by plugging in the x value of the specified point into your derivative, you'll get the SLOPE of the tangent line at that point. We'll call this result m.
Now, here's the point-slope form of a line, given point (x1, x2) and slope m
y-y1 = m(x - x1).
To get the point (x1, y1), we'll use the given point's x and y values, since we know these will be on the graph of the tangent line. If you're only given an x-value, plug it into the ORIGINAL equation to get the y-value.
So now you have an m, which you got by plugging the x-coordinate of the point into the derivative, and a point (x1, y1), which is just the point they gave you. This is the slope of the tangent line and the point at which the line is tangent to the graph, which is only significant because it guarantees this point will be on the tangent line.
y - y1 = m (x - x1)
Plug your values into the point-slope form of a line and simplify to get the equation of the tangent line.