I recognize you uncover the slope generally with \$dfracy_1-y_2x_1-x_2\$, however that doesn"t it seems ~ to job-related in this situation.

You are watching: How to calculate slope of a curved line  Write \$,Q=(5,8cdot b^5),,,,P=(1,8cdot b),\$ , climate the slope in between these two points is:

\$\$m_PQ=frac8b^5-8b5-1=2b(b^4-1)\$\$ It looks favor the original question had no calculus tag and also judging through the method the question was asked, I"m guessing no expertise of calculus.

The formula you posted because that slope works for a line whereby slope is constant. Without expertise of calculus, the ideal you can do is try to attract an approximate tangent line, the find the steep of the line.

This is a problem made for calculus though. The slope at any allude on a curve \$y=f(x)\$ is the derivative of \$y\$ with respect to \$x\$, written as \$fracdydx\$. Conan Wong provides the correct calculation for this derivative. \$\$b^x = e^ln (b^x) = e^x ln (b)\$\$

so utilizing the Chain Rule and also the reality that \$(e^x)" = e^x\$ we have

\$\$(b^x)" = (e^xln(b))" = (ln(b)) e^xln(b) = (ln (b))b^x\$\$

So the derivative that \$8b^x\$ is \$8(ln(b))b^x\$. Thanks because that contributing response to centregalilee.comematics Stack Exchange!

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