What is the tangent of y=ln(x^2),\at x=4 ?

The tangent of y=ln(x^2),\at x=4 is y= 1/2 x+4ln(2)-2

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tangent of y=ln(x^2),\at x=4

tangent of

y = ln (x^{2}), at x = 4

y = \frac{1}{2} x + 4 ln (2) - 2

Solution steps

Find the tangent point:

(4, 4 ln (2))

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 = 4 : m = \frac{1}{2}

Find the line with slope m=

\frac{1}{2}

and passing through

(4, 4 ln (2)) : y = \frac{1}{2} x + 4 ln (2) - 2

y = \frac{1}{2} x + 4 ln (2) - 2

\frac{\partial}{\partial y} (\frac{1}{x y ^{2} - x ^{2} y})

\int \frac{x}{3 + x ^{2}} d x

x y^{^{'}} - 2 y = x^{3}

y = x - 1, y = 0, x = 9

What is the tangent of y=ln(x^2),\at x=4 ?
The tangent of y=ln(x^2),\at x=4 is y= 1/2 x+4ln(2)-2