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

The tangent of f(x)=x^2sqrt(x^3+1),\at x=1 is y=(11)/(2sqrt(2))x+sqrt(2)-(11)/(2sqrt(2))

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tangent of f(x)=x^2sqrt(x^3+1),\at x=1

tangent of

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

y = \frac{11}{2 2} x + 2 - \frac{11}{2 2}

Solution steps

Find the tangent point:

(1, 2)

Find the slope of

f (x) = x^{2} x^{3} + 1 : \frac{df ( x )}{d x} = \frac{7 x ^{4} + 4 x}{2 x ^{3} + 1}

Compute the slope of

f (x) = x^{2} x^{3} + 1

x = 1 : m = \frac{11}{2 2}

Find the line with slope m=

\frac{11}{2 2}

and passing through

(1, 2) : y = \frac{11}{2 2} x + 2 - \frac{11}{2 2}

y = \frac{11}{2 2} x + 2 - \frac{11}{2 2}

x \to 1 lim (x^{2} + 3 x - 2)

\int_{0}^{1} x \cdot e^{- x} d x

\int_{0}^{1} (e^{- x}) d x

f (x) = x^{2} (1 - 9 x)

\int sin (3 x) cos (5 x) d x

What is the tangent of f(x)=x^2sqrt(x^3+1),\at x=1 ?
The tangent of f(x)=x^2sqrt(x^3+1),\at x=1 is y=(11)/(2sqrt(2))x+sqrt(2)-(11)/(2sqrt(2))