What is the tangent of y=x^4+5x^2-x,(1,5) ?

The tangent of y=x^4+5x^2-x,(1,5) is y=13x-8

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tangent of y=x^4+5x^2-x,(1,5)

tangent of

y = x^{4} + 5 x^{2} - x, (1, 5)

y = 13 x - 8

Solution steps

Find the slope of

y = x^{4} + 5 x^{2} - x : \frac{d y}{d x} = 4 x^{3} + 10 x - 1

Compute the slope of

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

(1, 5) : m = 13

Find the line with slope m=

13

and passing through

(1, 5) : y = 13 x - 8

y = 13 x - 8

t d t + y e^{- t} d y = 0, y (0) = 1

y^{^{''}} + 4 y^{^{'}} + 3 y = e^{x}

\frac{d}{d x} (x^{l n (x)})

\frac{d}{d x} (9 x^{2} - 2 x + 1)

x sin (\frac{4}{x})

What is the tangent of y=x^4+5x^2-x,(1,5) ?
The tangent of y=x^4+5x^2-x,(1,5) is y=13x-8