What is the tangent of f(x)=x^3,(1,1) ?

The tangent of f(x)=x^3,(1,1) is y=3x-2

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tangent of f(x)=x^3,(1,1)

tangent of

f (x) = x^{3}, (1, 1)

y = 3 x - 2

Solution steps

Find the slope of

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

Find the line with slope m=

3

and passing through

(1, 1) : y = 3 x - 2

y = 3 x - 2

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

\int \frac{5 x ^{2} + 11 x + 4}{x ^{3} + 3 x ^{2} + 3 x + 1} d x

\frac{d}{d x} (\frac{1}{( 4 x + 5 ) ^{2}})

\int \frac{x ln ( x ^{2} + 4 )}{x ^{2} + 4} d x

\int x^{2} x^{3} + 37 d x