What is the tangent of f(x)=9-ln(x),\at x=1 ?

The tangent of f(x)=9-ln(x),\at x=1 is y=-x+10

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tangent of f(x)=9-ln(x),\at x=1

tangent of

f (x) = 9 - ln (x), at x = 1

y = - x + 10

Solution steps

Find the tangent point:

(1, 9)

Find the slope of

f (x) = 9 - ln (x) : \frac{df ( x )}{d x} = - \frac{1}{x}

Compute the slope of

f (x) = 9 - ln (x)

x = 1 : m = - 1

Find the line with slope m=

- 1

and passing through

(1, 9) : y = - x + 10

y = - x + 10

\int \frac{1}{x 25 x ^{2} + 1} d x

f (x) = x^{- 3}

\frac{d}{d x} (5 sin (7 x))

\int (- y^{2} + 1) - (2 y^{2} - 2) d y

sec^{6} (x)

What is the tangent of f(x)=9-ln(x),\at x=1 ?
The tangent of f(x)=9-ln(x),\at x=1 is y=-x+10