What is the tangent of y=(x^3)/(x+3),(3, 9/2) ?

The tangent of y=(x^3)/(x+3),(3, 9/2) is y= 15/4 x-27/4

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tangent of y=(x^3)/(x+3),(3, 9/2)

tangent of

y = \frac{x ^{3}}{x + 3}, (3, \frac{9}{2})

y = \frac{15}{4} x - \frac{27}{4}

Solution steps

Find the slope of

y = \frac{x ^{3}}{x + 3} : \frac{d y}{d x} = \frac{2 x ^{3} + 9 x ^{2}}{( x + 3 ) ^{2}}

Compute the slope of

y = \frac{x ^{3}}{x + 3}

(3, \frac{9}{2}) : m = \frac{15}{4}

Find the line with slope m=

\frac{15}{4}

and passing through

(3, \frac{9}{2}) : y = \frac{15}{4} x - \frac{27}{4}

y = \frac{15}{4} x - \frac{27}{4}

\int \frac{2 cos ( x )}{sin ^{3} ( x )} d x

x cos (x - 3)

\int 2 x (x^{2} + 5)^{6} d x

\int \frac{1}{( x + y + z + 1 ) ^{3}} d z

x \to 0 lim (\frac{2. 7 ^{x} - 1}{x})

What is the tangent of y=(x^3)/(x+3),(3, 9/2) ?
The tangent of y=(x^3)/(x+3),(3, 9/2) is y= 15/4 x-27/4