What is the tangent of y=(x^4)+(3/((x^2))),\at x=-1 ?

The tangent of y=(x^4)+(3/((x^2))),\at x=-1 is y=2x+6

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tangent of y=(x^4)+(3/((x^2))),\at x=-1

tangent of

y = (x^{4}) + (\frac{3}{( x ^{2} )}), at x = - 1

y = 2 x + 6

Solution steps

Find the tangent point:

(- 1, 4)

Find the slope of

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

Compute the slope of

y = x^{4} + \frac{3}{x ^{2}}

x = - 1 : m = 2

Find the line with slope m=

2

and passing through

(- 1, 4) : y = 2 x + 6

y = 2 x + 6

y^{^{'}} + 2 y = t

e^{- a x^{2}}, x = 0

(x + 1) \cdot \frac{d y}{d x} + y = ln (x)

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

e^{, at} sin (b \cdot t)

What is the tangent of y=(x^4)+(3/((x^2))),\at x=-1 ?
The tangent of y=(x^4)+(3/((x^2))),\at x=-1 is y=2x+6