What is the tangent of x^3+7x-6 ?

The tangent of x^3+7x-6 is y=(3a_{0}^2+7)x-2a_{0}^3-6

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tangent of x^3+7x-6

tangent of

x^{3} + 7 x - 6

y = (3 a_{0}^{2} + 7) x - 2 a_{0}^{3} - 6

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{3} + 7 a_{0} - 6)

Find the slope of

y = x^{3} + 7 x - 6 : \frac{d y}{d x} = 3 x^{2} + 7

Compute the slope of

y = x^{3} + 7 x - 6

x = a_{0} : m = 3 a_{0}^{2} + 7

Find the line with slope m=

3 a_{0}^{2} + 7

and passing through

(a_{0}, a_{0}^{3} + 7 a_{0} - 6) : y = (3 a_{0}^{2} + 7) x - 2 a_{0}^{3} - 6

y = (3 a_{0}^{2} + 7) x - 2 a_{0}^{3} - 6

3 x

\int \frac{1}{z ^{2} - 1} d z

f (x) = (6 x^{2} - 10 x) e^{x}

\int (sec^{2} (θ) + 5) d θ

\frac{d y}{d x} + y^{9} x + 7 y = 0

What is the tangent of x^3+7x-6 ?
The tangent of x^3+7x-6 is y=(3a_{0}^2+7)x-2a_{0}^3-6