What is the tangent of y=sqrt(x^3+1),(2,3) ?

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

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tangent of y=sqrt(x^3+1),(2,3)

tangent of

y = x^{3} + 1, (2, 3)

y = 2 x - 1

Solution steps

Find the slope of

y = x^{3} + 1 : \frac{d y}{d x} = \frac{3 x ^{2}}{2 x ^{3} + 1}

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 = 2

Find the line with slope m=

2

and passing through

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

y = 2 x - 1

What is the tangent of y=sqrt(x^3+1),(2,3) ?
The tangent of y=sqrt(x^3+1),(2,3) is y=2x-1