What is the tangent of f(x)=7ln(x),\at x=3 ?

The tangent of f(x)=7ln(x),\at x=3 is y= 7/3 x+7ln(3)-7

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tangent of f(x)=7ln(x),\at x=3

tangent of

f (x) = 7 ln (x), at x = 3

y = \frac{7}{3} x + 7 ln (3) - 7

Solution steps

Find the tangent point:

(3, 7 ln (3))

Find the slope of

f (x) = 7 ln (x) : \frac{df ( x )}{d x} = \frac{7}{x}

Compute the slope of

f (x) = 7 ln (x)

x = 3 : m = \frac{7}{3}

Find the line with slope m=

\frac{7}{3}

and passing through

(3, 7 ln (3)) : y = \frac{7}{3} x + 7 ln (3) - 7

y = \frac{7}{3} x + 7 ln (3) - 7

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

f (x) = (2 x + 1) (3 x - 2)

\int_{3}^{e^{4}} \frac{1}{x} d x

y = ln (x), y = 0, x = e

\int cos (2) (x) d x

What is the tangent of f(x)=7ln(x),\at x=3 ?
The tangent of f(x)=7ln(x),\at x=3 is y= 7/3 x+7ln(3)-7