What is the tangent of f(x)= 1/2 xln(x^2),(-1,0) ?

The tangent of f(x)= 1/2 xln(x^2),(-1,0) is y=x+1

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tangent of f(x)= 1/2 xln(x^2),(-1,0)

tangent of

f (x) = \frac{1}{2} x ln (x^{2}), (- 1, 0)

y = x + 1

Solution steps

Find the slope of

f (x) = \frac{1}{2} x ln (x^{2}) : \frac{df ( x )}{d x} = \frac{ln ( x ^{2} ) + 2}{2}

Compute the slope of

f (x) = \frac{1}{2} x ln (x^{2})

(- 1, 0) : m = 1

Find the line with slope m=

1

and passing through

(- 1, 0) : y = x + 1

y = x + 1

\int cos (θ) + sin (θ) d θ

\frac{d}{d x} (\frac{2 x}{x ^{2} + 9})

\int_{0}^{1} x sin (\frac{nπ x}{2}) d x

x \to y lim (x - y)

x \to - 1 lim (2 - x)

What is the tangent of f(x)= 1/2 xln(x^2),(-1,0) ?
The tangent of f(x)= 1/2 xln(x^2),(-1,0) is y=x+1