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Find an equation of the tangent line to the curve y=8^x at the point (2,64) Tangent line y=
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Find the first derivative to find the slope at that point. y'=ln(8) * 8^x
once you have the slope you can find the rest of the equation using y=mx+b. Use your point for y and x and solve or b
So would I plug in 64 for y and 2 for x into the first derivative to get the slope? Which would make it 2.0794....5?
the first derivative can also be written like this \[dy/dt=\ln(8)8^x\] use the x value to find the slope.
dy/dx sorry
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the slope would be 133.08426
alright so I got 133.084=ln(8)*8^2 I now go 64=133.084(2)+b to find b?
you got it!
Thank You so much. I appreciate it
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