What is the tangent of f(x)=(x^3)/(x+1),\at x=1 ?

The tangent of f(x)=(x^3)/(x+1),\at x=1 is y= 5/4 x-3/4

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tangent of f(x)=(x^3)/(x+1),\at x=1

tangent of

f (x) = \frac{x ^{3}}{x + 1}, at x = 1

y = \frac{5}{4} x - \frac{3}{4}

Solution steps

Find the tangent point:

(1, \frac{1}{2})

Find the slope of

f (x) = \frac{x ^{3}}{x + 1} : \frac{df ( x )}{d x} = \frac{2 x ^{3} + 3 x ^{2}}{( x + 1 ) ^{2}}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = \frac{5}{4}

Find the line with slope m=

\frac{5}{4}

and passing through

(1, \frac{1}{2}) : y = \frac{5}{4} x - \frac{3}{4}

y = \frac{5}{4} x - \frac{3}{4}

\frac{d}{d x} (tan (π x - 1) + e^{c o s (x)})

y = (x - 1)^{3}

\int tan (x) ln (cos (x)) d x

\int \frac{x ^{2}}{x ^{2} + 4} d x

y = x^{2} - 1, y = 2 x - 1, y = x + 1

What is the tangent of f(x)=(x^3)/(x+1),\at x=1 ?
The tangent of f(x)=(x^3)/(x+1),\at x=1 is y= 5/4 x-3/4