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

The tangent of-2x^5+4x^4 is y=(-10a_{0}^4+16a_{0}^3)x+8a_{0}^5-12a_{0}^4

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

tangent of

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

y = (- 10 a_{0}^{4} + 16 a_{0}^{3}) x + 8 a_{0}^{5} - 12 a_{0}^{4}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

y = - 2 x^{5} + 4 x^{4} : \frac{d y}{d x} = - 10 x^{4} + 16 x^{3}

Compute the slope of

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

x = a_{0} : m = - 10 a_{0}^{4} + 16 a_{0}^{3}

Find the line with slope m=

- 10 a_{0}^{4} + 16 a_{0}^{3}

and passing through

(a_{0}, - 2 a_{0}^{5} + 4 a_{0}^{4}) : y = (- 10 a_{0}^{4} + 16 a_{0}^{3}) x + 8 a_{0}^{5} - 12 a_{0}^{4}

y = (- 10 a_{0}^{4} + 16 a_{0}^{3}) x + 8 a_{0}^{5} - 12 a_{0}^{4}

\frac{d}{d x} (\frac{12}{x} - \frac{4}{x ^{3}} + \frac{1}{x ^{4}})

y^{^{''}} + 2 y^{^{'}} + y = 0, y (2) = 1, y^{^{'}} (2) = - 2

\int \frac{( x ^{2} - x + 1 )}{( x ^{2} + x )} d x

\frac{d}{d x} (11 x - \frac{6}{x})

\int t^{6} e^{- t^{7}} d t

What is the tangent of-2x^5+4x^4 ?
The tangent of-2x^5+4x^4 is y=(-10a_{0}^4+16a_{0}^3)x+8a_{0}^5-12a_{0}^4