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}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 10

Find the line with slope m=

10

and passing through

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

y = 10 x - 2

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