What is the tangent of f(x)=6+ln(x),\at x=1 ?

The tangent of f(x)=6+ln(x),\at x=1 is y=x+5

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tangent of f(x)=6+ln(x),\at x=1

tangent of

f (x) = 6 + ln (x), at x = 1

y = x + 5

Solution steps

Find the tangent point:

(1, 6)

Find the slope of

f (x) = 6 + ln (x) : \frac{df ( x )}{d x} = \frac{1}{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 = 1

Find the line with slope m=

1

and passing through

(1, 6) : y = x + 5

y = x + 5

\frac{\partial}{\partial x} (e^{3 x} (x^{2} + y^{2}))

e^{- 5 x} + e^{2 x}

x \to - 2 lim (x^{2} - x)

\frac{d}{d x} (x \cdot sin (x) + cos (x))

\frac{20}{s ^{2} + 10}

What is the tangent of f(x)=6+ln(x),\at x=1 ?
The tangent of f(x)=6+ln(x),\at x=1 is y=x+5