What is the tangent of f(x)=8x+ln(x^2),\at x=1 ?

The tangent of f(x)=8x+ln(x^2),\at x=1 is y=10x-2

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tangent of f(x)=8x+ln(x^2),\at x=1

tangent of

f (x) = 8 x + ln (x^{2}), at x = 1

y = 10 x - 2

Solution steps

Find the tangent point:

(1, 8)

Find the slope of

f (x) = 8 x + ln (x^{2}) : \frac{df ( x )}{d x} = 8 + \frac{2}{x}

Compute the slope of

f (x) = 8 x + ln (x^{2})

x = 1 : m = 10

Find the line with slope m=

10

and passing through

(1, 8) : y = 10 x - 2

y = 10 x - 2

\int \frac{x ^{6} - x ^{3} + 36}{x ^{4} + 6 x ^{2}} d x

\frac{d}{d x} (\frac{sin ( x )}{e ^{x}})

\int \frac{x}{2 x - 3} d x

\frac{\partial}{\partial x} (2 x y^{3} - 1)

\frac{d y}{d t} = \frac{t}{y + 1}

What is the tangent of f(x)=8x+ln(x^2),\at x=1 ?
The tangent of f(x)=8x+ln(x^2),\at x=1 is y=10x-2