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Consider the curve y=ln(3x-1) Let P be the point on the curve where x = 2.
(a) Write down the gradient of the curve at P.

(the anser is 0.6 but idk how they got there)

Respuesta :

[tex]y = \ln(3x-1) \\ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{3x-1}\cdot 3 = \frac{3}{3x-1}[/tex]

[tex]x=2[/tex], so the gradient of this curve at [tex]P[/tex] is [tex]\frac{3}{3\cdot 2-1} = \frac{3}{6-1} = \frac{3}{5} = 0.6[/tex].

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