What is the tangent of (x^3+2)^5,\at x=-1 ?

The tangent of (x^3+2)^5,\at x=-1 is y=15x+16

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tangent of (x^3+2)^5,\at x=-1

tangent of

(x^{3} + 2)^{5}, at x = - 1

y = 15 x + 16

Solution steps

Find the tangent point:

(- 1, 1)

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 15

Find the line with slope m=

15

and passing through

(- 1, 1) : y = 15 x + 16

y = 15 x + 16

\frac{\partial}{\partial x} (56 y^{4} x^{6} - 42 y^{3} x^{5})

(x - 1)^{2} (x - 2)^{3}

n = 1 \sum \infty 8 ((\frac{1}{5})^{n})

x \to 0 lim (\frac{6}{x} - \frac{6}{x ^{2} - x})

\int_{1}^{9} \frac{1}{9 - x} d x

What is the tangent of (x^3+2)^5,\at x=-1 ?
The tangent of (x^3+2)^5,\at x=-1 is y=15x+16