What is the tangent of y=4ln(x),\at x=4 ?

The tangent of y=4ln(x),\at x=4 is y=x+8ln(2)-4

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tangent of y=4ln(x),\at x=4

tangent of

y = 4 ln (x), at x = 4

y = x + 8 ln (2) - 4

Solution steps

Find the tangent point:

(4, 8 ln (2))

Find the slope of

y = 4 ln (x) : \frac{d y}{d x} = \frac{4}{x}

Compute the slope of

y = 4 ln (x)

x = 4 : m = 1

Find the line with slope m=

1

and passing through

(4, 8 ln (2)) : y = x + 8 ln (2) - 4

y = x + 8 ln (2) - 4

\int_{0}^{1} (x^{\frac{4}{5}}) d x

\int \frac{( 1 + 6 ln ( x ) ) ^{11}}{x} d x

\int x^{2} + 2 x y + y^{2} d x

\frac{d y}{d x} = \frac{1}{1 - x ^{2}} - \frac{1}{x}

x^{2} + 4 x + 6

What is the tangent of y=4ln(x),\at x=4 ?
The tangent of y=4ln(x),\at x=4 is y=x+8ln(2)-4