What is the tangent of f(x)=x^3ln(2x),\at x=5 ?

The tangent of f(x)=x^3ln(2x),\at x=5 is y=(75ln(10)+25)x-250ln(10)-125

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tangent of f(x)=x^3ln(2x),\at x=5

tangent of

f (x) = x^{3} ln (2 x), at x = 5

y = (75 ln (10) + 25) x - 250 ln (10) - 125

Solution steps

Find the tangent point:

(5, 125 ln (10))

Find the slope of

f (x) = x^{3} ln (2 x) : \frac{df ( x )}{d x} = 3 x^{2} ln (2 x) + x^{2}

Compute the slope of

f (x) = x^{3} ln (2 x)

x = 5 : m = 75 ln (10) + 25

Find the line with slope m=

75 ln (10) + 25

and passing through

(5, 125 ln (10)) : y = (75 ln (10) + 25) x - 250 ln (10) - 125

y = (75 ln (10) + 25) x - 250 ln (10) - 125

x \to 1 lim (1 - (x - 1)^{2})

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

\int e^{x} cos (2 x) d x

\frac{d y}{d x} = k y^{2} ln (x), y (1) = - 5

(x^{3} - y^{3}) d x + x y^{2} d y = 0

What is the tangent of f(x)=x^3ln(2x),\at x=5 ?
The tangent of f(x)=x^3ln(2x),\at x=5 is y=(75ln(10)+25)x-250ln(10)-125