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

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

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

tangent of

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

y = 13 x - 8

Solution steps

Find the slope of

f (x) = x^{4} + 5 x^{2} - x : \frac{df ( x )}{d x} = 4 x^{3} + 10 x - 1

Compute the slope of

f (x) = 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

\int \frac{4 x + 3}{x ^{2} + 4} d x

(x^{4} - x + y) d x - x d y = 0

ln (x x^{2} + 1)

f (x) = 6 x - 6 x

\int_{1}^{2} x e^{x^{2}} d x

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