<p>Hey, I have a test tomorrow on implict and explicit differentation-it is pretty simple, but one question is stumping me:</p>
<p>Find an equation for a line that is tangent to the grap of y=e^x and goes through the origin.</p>
<p>The derivative is obviously y'=e^x, but wouldn't the slope at the origin simply be 1? The answer in the book is y=ex, but I don't get how the slope can be e.</p>
<p>Am I misunderstanding the question in any way?</p>
<p>Bump Bump...</p>
<p>Careful. The slope of y=e^x is itself e^x, and its value changes with x. At x=0, the slope of e^x is indeed 1. However, the straight line (say g(x)) = kx, and has a slope of k.</p>
<p>You need to solve for x where
1) f(x) = g(x) , or e^x = kx
and 2) the slope of f(x) equals the slope of the straight line, or e^x = k</p>
<p>From these two conditions, you get kx = e^x = k
If kx = k, then x = 1.
And since e^x must = kx, e^1 = (k)(1), or k = e.</p>
<p>Ohhhhhh.....if it passes through the origin at the graph of e^x, is must be 1 because the tangent line must have a slope of 1, so you plug it in to get e.</p>
<p>Thanks.</p>
<p>Good. Now tell me how you'd solve the problem if y = e^2x :)</p>
<p>Well the derivative is 2e^2x.</p>
<p>The slope: (2e^2x-0)/(x-0)=2e^2x/x</p>
<p>(2e^2x/x)=e^2x</p>
<p>x=2 so the equation: y=(2e^4)(x)</p>
<p>Are you sure?</p>
<p>You need to solve for x where
1) f(x) = g(x) , or e^2x = kx
2) the slope of f(x) equals the slope of the straight line, or
2e^2x = k</p>
<p>From these two conditions, you get kx = e^2x = k/2
If kx = k/2, then x = 1/2.
And since e^2x must = kx, e^1 = (k)(1/2), or k = 2e.</p>
<p>So the straight line has the equation g(x) = (2e)x, and it touches f(x) at (0.5, e) .</p>
<p>[ Be careful - you correctly had f'(x) = 2e^2x, but you are mixing it up with the limits expression used to define derivatives]</p>