Can any of you guys walk me through what to do on #16? I know how to find the derivative, but am clueless as to how I would relate it to finding a tangent line.
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Can any of you guys walk me through what to do on #16? I know how to find the derivative, but am clueless as to how I would relate it to finding a tangent line.
Please keep in mind that the derivative is the equation for the slope of the tangent line. To form the equation of the line you need a point and a slope.
It's asking you to find the equation of the tangent line given an equation and a point.
Start off by finding out what the point is by plugging in x into the equation.
x = 0
y = e^(0+1) y = 2.72
pt = (0, 2.72)
y' = e^(x+1)
Remembering that the derivative is the slope of the tangent line allows us to plug x into the derivative to get the slope of the tangent line at that point.
f'(0) = 2.72
m = 2.72
Now, using the point slope formula you can create an equation for the tangent line:
y-y1= m(x-x1)
where y1 and x1 are the point you found and m is the slope you calculated.
y-2.72 = 2.72(x-0)
y-2.72 = 2.72x
y = 2.72x + 2.72
Please correct me if I went wrong anywhere.
EDIT: Please note 2.72 = e^1
An easy way to remember is to just think of the derivative as a way of saying the "rate of change" of a function. If you made a bunch of tangent lines on your original function and proceeded to calculate their slopes, the same effect can be modelled by taking the derivative and producing a general equation to represent this procedure/act. Remembering the non calculus equivalent and the true definition of a derivative (have you done this yet?) helps make things intuitive.
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