What is the tangent of-3(x)^5+9(x)^3,\at x=-1 ?

The tangent of-3(x)^5+9(x)^3,\at x=-1 is y=12x+6

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tangent of-3(x)^5+9(x)^3,\at x=-1

tangent of

- 3 (x)^{5} + 9 (x)^{3}, at x = - 1

y = 12 x + 6

Solution steps

Find the tangent point:

(- 1, - 6)

Find the slope of

y = - 3 x^{5} + 9 x^{3} : \frac{d y}{d x} = - 15 x^{4} + 27 x^{2}

Compute the slope of

y = - 3 x^{5} + 9 x^{3}

x = - 1 : m = 12

Find the line with slope m=

12

and passing through

(- 1, - 6) : y = 12 x + 6

y = 12 x + 6

4 x^{2}, 16 x

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

\frac{d}{d x} (x^{0.5} \cdot y^{0.5})

f (x) = e^{9 x}

\frac{d x}{d t} = \frac{2500 \cdot 7.75}{( 1500 - 7.75 t )}

What is the tangent of-3(x)^5+9(x)^3,\at x=-1 ?
The tangent of-3(x)^5+9(x)^3,\at x=-1 is y=12x+6