Find d/dx (ln(2x^3+x+8))
Use the chain rule: dy/dx = dy/du * du/dx
Let y = ln(2x^3+x+8)Let u = 2x^3+x+8
dy/dx = d/dx (ln(2x^3+x+8)) = dy/du * du/dx
dy/du = 1/udu/dx = 6x^2 + 1
dy/dx = 1/u * (6x^2 + 1) = (1/(2x^3+x+8)) * (6x^2 + 1) = (6x^2 + 1) / (2x^3+x+8)
d/dx (ln(2x^3+x+8)) = (6x^2 + 1) / (2x^3+x+8)
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