need to find the slope of the tangent line to the curve y=lnx^8 at the point where x=e^2
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Originally Posted by pebblesbambam need to find the slope of the tangent line to the curve y=lnx^8 at the point where x=e^2 $\displaystyle y = \ln{x^8}$ $\displaystyle y = 8\ln{x} $ $\displaystyle y' =$ ? once you determine the derivative, evaluate at $\displaystyle x = e^2$
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