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

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

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

tangent of

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

y = (6 a_{0} (a_{0}^{2} + 1)^{2} (a_{0} - 1) + \frac{( a _{0}^{2} + 1 ) ^{3}}{2 a _{0}}) x - 5 a_{0}^{6} a_{0} + 5 a_{0}^{6} - 9 a_{0}^{4} a_{0} + 9 a_{0}^{4} - 3 a_{0}^{2} a_{0} + 3 a_{0}^{2} + a_{0} - \frac{a _{0} ( a _{0}^{2} + 1 ) ^{3}}{2} - 1

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, (a_{0}^{2} + 1)^{3} (a_{0} - 1))

Find the slope of

y = (x^{2} + 1)^{3} (x - 1) : \frac{d y}{d x} = 6 x (x^{2} + 1)^{2} (x - 1) + \frac{( x ^{2} + 1 ) ^{3}}{2 x}

Compute the slope of

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

x = a_{0} : m = 6 a_{0} (a_{0}^{2} + 1)^{2} (a_{0} - 1) + \frac{( a _{0}^{2} + 1 ) ^{3}}{2 a _{0}}

Find the line with slope m=

6 a_{0} (a_{0}^{2} + 1)^{2} (a_{0} - 1) + \frac{( a _{0}^{2} + 1 ) ^{3}}{2 a _{0}}

and passing through

(a_{0}, (a_{0}^{2} + 1)^{3} (a_{0} - 1)) : y = (6 a_{0} (a_{0}^{2} + 1)^{2} (a_{0} - 1) + \frac{( a _{0}^{2} + 1 ) ^{3}}{2 a _{0}}) x - 5 a_{0}^{6} a_{0} + 5 a_{0}^{6} - 9 a_{0}^{4} a_{0} + 9 a_{0}^{4} - 3 a_{0}^{2} a_{0} + 3 a_{0}^{2} + a_{0} - \frac{a _{0} ( a _{0}^{2} + 1 ) ^{3}}{2} - 1

y = (6 a_{0} (a_{0}^{2} + 1)^{2} (a_{0} - 1) + \frac{( a _{0}^{2} + 1 ) ^{3}}{2 a _{0}}) x - 5 a_{0}^{6} a_{0} + 5 a_{0}^{6} - 9 a_{0}^{4} a_{0} + 9 a_{0}^{4} - 3 a_{0}^{2} a_{0} + 3 a_{0}^{2} + a_{0} - \frac{a _{0} ( a _{0}^{2} + 1 ) ^{3}}{2} - 1

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