What is the tangent of y=(x^2-3)(x^3-2x),(-2,-4) ?

The tangent of y=(x^2-3)(x^3-2x),(-2,-4) is y=26x+48

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tangent of y=(x^2-3)(x^3-2x),(-2,-4)

tangent of

y = (x^{2} - 3) (x^{3} - 2 x), (- 2, - 4)

y = 26 x + 48

Solution steps

Find the slope of

y = (x^{2} - 3) (x^{3} - 2 x) : \frac{d y}{d x} = 5 x^{4} - 15 x^{2} + 6

Compute the slope of

y = (x^{2} - 3) (x^{3} - 2 x)

(- 2, - 4) : m = 26

Find the line with slope m=

26

and passing through

(- 2, - 4) : y = 26 x + 48

y = 26 x + 48

\frac{d}{d x} (\frac{lo g _{7} ( x - x ^{2} )}{x ^{3}})

[x^{2} (x^{2} + 5 x)]^{4}

e^{x^{4}}

x \to 3 lim ((x - 1)^{2})

\frac{d ^{2} y}{d x ^{2}} - 3 \frac{d y}{d x} + 2 y = 0

What is the tangent of y=(x^2-3)(x^3-2x),(-2,-4) ?
The tangent of y=(x^2-3)(x^3-2x),(-2,-4) is y=26x+48