What is the tangent of y=ln(x^2),\at x=3 ?

The tangent of y=ln(x^2),\at x=3 is y= 2/3 x+2ln(3)-2

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tangent of y=ln(x^2),\at x=3

tangent of

y = ln (x^{2}), at x = 3

y = \frac{2}{3} x + 2 ln (3) - 2

Solution steps

Find the tangent point:

(3, 2 ln (3))

Find the slope of

y = ln (x^{2}) : \frac{d y}{d x} = \frac{2}{x}

Compute the slope of

y = ln (x^{2})

x = 3 : m = \frac{2}{3}

Find the line with slope m=

\frac{2}{3}

and passing through

(3, 2 ln (3)) : y = \frac{2}{3} x + 2 ln (3) - 2

y = \frac{2}{3} x + 2 ln (3) - 2

\int sinh^{2} (x) d x

\int_{0}^{4} x (x + 2) d x

x^{^{''}} + 5 x^{^{'}} = 7

\int \frac{4}{t + 1} d t

\frac{d}{d x} (x^{6} e^{8 x})

What is the tangent of y=ln(x^2),\at x=3 ?
The tangent of y=ln(x^2),\at x=3 is y= 2/3 x+2ln(3)-2