gradient of the curve at that point. You find the gradient by taking
the derivative of the equation of the curve, so:
dy/dx=4* 1/2 * 1/sqrt(x)
(note I am using "*" as the multiplication symbol here, which is a common
convention when writting maths in plain ASCII)
At the point in question x=7, so the gradient is:
dy/dx = 2/sqrt(7)~=0.75593.
Now we know the slope of the line we can write the equation of the
line in the form:
but we also know that this line is a tangent to the curve at (7,10.5831),
so we know that:
Which can be solved for b to give the full equation of the desired line.