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

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

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

tangent of

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

y = 15 x + 16

Solution steps

Find the tangent point:

(- 1, 1)

Find the slope of

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

Compute the slope of

f (x) = (x^{3} + 2)^{5}

x = - 1 : m = 15

Find the line with slope m=

15

and passing through

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

y = 15 x + 16

x, 1

\frac{d}{d x} (4 - 3 x^{2})

y = x^{3} - 6 x^{2} + 8 x, y = x^{2} - 4 x

ln (x e^{x} + 2)

\frac{d}{d x} (\frac{2 cot ( x )}{csc ( x )})

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