What is the tangent of f(x)=x^3,(2,8) ?

The tangent of f(x)=x^3,(2,8) is y=12x-16

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tangent of f(x)=x^3,(2,8)

tangent of

f (x) = x^{3}, (2, 8)

y = 12 x - 16

Solution steps

Find the slope of

f (x) = x^{3} : \frac{df ( x )}{d x} = 3 x^{2}

Compute the slope of

f (x) = x^{3}

(2, 8) : m = 12

Find the line with slope m=

12

and passing through

(2, 8) : y = 12 x - 16

y = 12 x - 16

\frac{d}{d x} (\frac{5}{2 x + 1})

f (x) = 3 sin (x) + 2 cos (x)

\frac{d}{d x} (sinh^{3} (2 x))

\int \frac{2 x ^{2} - 25 x - 33}{( x + 1 ) ^{2} ( x - 5 )} d x

3 e^{5 x} \frac{d y}{d x} = \frac{- 25 x}{y ^{2}}, y (0) = 1

What is the tangent of f(x)=x^3,(2,8) ?
The tangent of f(x)=x^3,(2,8) is y=12x-16