I understand you find the slope commonly with \$dfracy_1-y_2x_1-x_2\$, but that doesn"t it seems to be ~ to work in this situation.

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

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

The formula friend posted for slope functions for a line wherein slope is constant. Without expertise of calculus, the finest you deserve to do is shot to draw an approximate tangent line, the uncover the slope of the line.

This is a trouble made for calculus though. The slope in ~ any suggest on a curve \$y=f(x)\$ is the derivative the \$y\$ with respect come \$x\$, written as \$fracdydx\$. Conan Wong offers the correct calculation because that this derivative. \$\$b^x = e^ln (b^x) = e^x ln (b)\$\$

so making use of the Chain Rule and the reality that \$(e^x)" = e^x\$ we have actually

\$\$(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 6294.orgematics Stack Exchange!

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