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$f(x) = 8 b^x$

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My answer will be in terms of $b$.

I know you find the slope typically with $\dfrac{y_1-y_2}{x_1-x_2}$, but that doesn't seem to work in this situation.

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    The slope of a smooth curve at a pont on the curve is defined as the slope of the tangent line at that point. Calculation of this requires some concepts from the calculus. But finding the slope of the *straight* line that joins $p$ and $q$ just requires the formula you quoted. Here $x_1$ and $x_2$ are given to you, but you need to calculate $y_1$ and $y_2$.2012-11-27

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Write $\,Q=(5,8\cdot b^5)\,\,,\,P=(1,8\cdot b)\,$ , then the slope between these two points is:

$m_{PQ}=\frac{8b^5-8b}{5-1}=2b(b^4-1)$

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    Well, that's what the function is, not my call! If we have a function $\,y=f(x)\,$, then a point on the plane belongs to the function's graph iff the point has the form $\,(x_0,f(x_0))\,$ , for some number $\,x_0\,$ in the function's domain. This is what I applied here.2012-11-27
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It looks like the original question had no calculus tag and judging by the way the question was asked, I'm guessing no knowledge of calculus.

The formula you posted for slope works for a line where slope is constant. Without knowledge of calculus, the best you can do is try to draw an approximate tangent line, the find the slope of the line.

This is a problem made for calculus though. The slope at any point on a curve $y=f(x)$ is the derivative of $y$ with respect to $x$, written as $\frac{dy}{dx}$. Conan Wong gives the correct calculation for this derivative.

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    @DonAntonio It isn't quite clear from the question, but had I noticed Andre's comment, I probably would have upvoted that rather than giving this answer as it seems he had it all covered.2012-11-27
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$b^x = e^{ln (b^x)} = e^{x ln (b)}$

so using the Chain Rule and the fact that $(e^x)' = e^x$ we have

$(b^x)' = (e^{xln(b)})' = (ln(b)) e^{xln(b)} = (ln (b))b^x$

So the derivative of $8b^x$ is $8(ln(b))b^x$.

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    @DonAntonio, yeah, I only figured that out after re-reading Tyler's question. But anyway, good practice for me :-)2012-11-27