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

The tangent of y=4ln(x) is y= 4/(a_{0)}x+4ln(a_{0})-4

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tangent of y=4ln(x)

tangent of

y = 4 ln (x)

y = \frac{4}{a _{0}} x + 4 ln (a_{0}) - 4

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 4 ln (a_{0}))

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 = a_{0} : m = \frac{4}{a _{0}}

Find the line with slope m=

\frac{4}{a _{0}}

and passing through

(a_{0}, 4 ln (a_{0})) : y = \frac{4}{a _{0}} x + 4 ln (a_{0}) - 4

y = \frac{4}{a _{0}} x + 4 ln (a_{0}) - 4

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

\int \frac{4}{x - 2} d x

\int 3 (3 x + 4)^{5} d x

\int \frac{sin ( 2 x )}{45 + cos ^{2} ( x )} d x

\frac{d}{d x} (1 - 4 x x^{3})

What is the tangent of y=4ln(x) ?
The tangent of y=4ln(x) is y= 4/(a_{0)}x+4ln(a_{0})-4