Question

find the equation of the tangent line of y = 3x2 − ln(x) for (1,3)

Answer 1

Not sure how to set this up.

Answer 2

Calculus is needed here, are you up for it?

Answer 3

yeah I'm always ready :P

Answer 4

Ok, first step is to find dy/dx, this is the slope of the tangent line at any given point

Answer 5

It says the answer is 5x-2. So I figured deriving would be 6x - 1/x

Answer 6

plugging in x would be 6* 1 - 1/1 = 5

Answer 7

yeah, so 5 is the slope of the line
and we have a point (1,3)
so all that's left is to determine the equation of the line
from the point and the slope

Answer 8

so you get
y - 3 = m(x - 1)
y - 3 = 5x - 5
y = 5x - 2
hmm I'm slow haha

Answer 9

Complete step-by-step

slope at x=1 (given by (1,3)
\[\frac{d}{dx} y = \frac{d}{dx} 3x^2 - \frac{d}{dx} \ln(x)\]
applying power rule and chain rule
\[\frac{d}{dx} y = \frac{d}{dx} (3)(2)x^{2-1} - \frac{d}{dx} \frac{1}{ln(e)*x}\]
\[\frac{dy}{dx} = 6x - \frac{1}{x}\]
\[\frac{dy}{dx} = 6(1) - \frac{1}{(1)}=5\]

\[y - y_0 = m(x - x_0) \]
\[y - 3 = m(x - 1) \]
\[y - 3 = 5x - 5 \]
\[y = 5x - 2\]

Answer 10

Sorry for not replying, my internet dropped again :(

Answer 11

wait the chain rule is in there @ agent5x?