What is the tangent of y=x^3-7x^2+16x+7,\at ?

The tangent of y=x^3-7x^2+16x+7,\at is y=(3a_{0}^2-14a_{0}+16)x-2a_{0}^3+7a_{0}^2+7

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tangent of y=x^3-7x^2+16x+7,\at

tangent of

y = x^{3} - 7 x^{2} + 16 x + 7, at

y = (3 a_{0}^{2} - 14 a_{0} + 16) x - 2 a_{0}^{3} + 7 a_{0}^{2} + 7

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}^{2} + 16 a_{0} + 7)

Find the slope of

y = x^{3} - 7 x^{2} + 16 x + 7 : \frac{d y}{d x} = 3 x^{2} - 14 x + 16

Compute the slope of

y = x^{3} - 7 x^{2} + 16 x + 7

x = a_{0} : m = 3 a_{0}^{2} - 14 a_{0} + 16

Find the line with slope m=

3 a_{0}^{2} - 14 a_{0} + 16

and passing through

(a_{0}, a_{0}^{3} - 7 a_{0}^{2} + 16 a_{0} + 7) : y = (3 a_{0}^{2} - 14 a_{0} + 16) x - 2 a_{0}^{3} + 7 a_{0}^{2} + 7

y = (3 a_{0}^{2} - 14 a_{0} + 16) x - 2 a_{0}^{3} + 7 a_{0}^{2} + 7

\frac{d}{d x} (\frac{6 x}{7 - tan ( x )})

(- 8 cos (2 x))^{^{'}}

f (x) = 1 + x

\frac{d}{d x} (x (3 x + 2)^{4})

\frac{\partial}{\partial x} (9 x ln (x) y)

What is the tangent of y=x^3-7x^2+16x+7,\at ?
The tangent of y=x^3-7x^2+16x+7,\at is y=(3a_{0}^2-14a_{0}+16)x-2a_{0}^3+7a_{0}^2+7