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

The tangent of f(x)=sqrt(3x^3),\at x=3 is y= 9/2 x-9/2

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tangent of f(x)=sqrt(3x^3),\at x=3

tangent of

f (x) = 3 x^{3}, at x = 3

y = \frac{9}{2} x - \frac{9}{2}

Solution steps

Find the tangent point:

(3, 9)

Find the slope of

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

Compute the slope of

f (x) = 3 x^{3}

x = 3 : m = \frac{9}{2}

Find the line with slope m=

\frac{9}{2}

and passing through

(3, 9) : y = \frac{9}{2} x - \frac{9}{2}

y = \frac{9}{2} x - \frac{9}{2}

\frac{\partial}{\partial x} (x^{2} e^{y} + x - 2 y)

\int \frac{x ^{2} + 2 x + 1}{x ^{3} + x} d x

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

\int 169 - t^{2} d t

x \frac{d y}{d x} + y = 6 x + 1

What is the tangent of f(x)=sqrt(3x^3),\at x=3 ?
The tangent of f(x)=sqrt(3x^3),\at x=3 is y= 9/2 x-9/2