What is the tangent of x^4-2x^3+5 ?

The tangent of x^4-2x^3+5 is y=(4a_{0}^3-6a_{0}^2)x-3a_{0}^4+4a_{0}^3+5

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tangent of x^4-2x^3+5

tangent of

x^{4} - 2 x^{3} + 5

y = (4 a_{0}^{3} - 6 a_{0}^{2}) x - 3 a_{0}^{4} + 4 a_{0}^{3} + 5

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{4} - 2 a_{0}^{3} + 5)

Find the slope of

y = x^{4} - 2 x^{3} + 5 : \frac{d y}{d x} = 4 x^{3} - 6 x^{2}

Compute the slope of

y = x^{4} - 2 x^{3} + 5

x = a_{0} : m = 4 a_{0}^{3} - 6 a_{0}^{2}

Find the line with slope m=

4 a_{0}^{3} - 6 a_{0}^{2}

and passing through

(a_{0}, a_{0}^{4} - 2 a_{0}^{3} + 5) : y = (4 a_{0}^{3} - 6 a_{0}^{2}) x - 3 a_{0}^{4} + 4 a_{0}^{3} + 5

y = (4 a_{0}^{3} - 6 a_{0}^{2}) x - 3 a_{0}^{4} + 4 a_{0}^{3} + 5

\int_{0}^{t} x d x

\frac{d}{d x} (cos (3 x^{7} + 2))

\int sec (4 x) d x

\frac{d}{d x} (a cos (x) - b sin (x))

\frac{d}{d x} (\frac{24}{( x - 1 ) ^{5}})

What is the tangent of x^4-2x^3+5 ?
The tangent of x^4-2x^3+5 is y=(4a_{0}^3-6a_{0}^2)x-3a_{0}^4+4a_{0}^3+5