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

The tangent of f(x)=5x^3,\at x=2 is y=60x-80

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

tangent of

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

y = 60 x - 80

Solution steps

Find the tangent point:

(2, 40)

Find the slope of

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

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 = 60

Find the line with slope m=

60

and passing through

(2, 40) : y = 60 x - 80

y = 60 x - 80

\frac{\partial}{\partial x} (x y + cos (x))

\frac{d}{d x} f (x) = (2 x - 6)^{2} (x^{2} + x + 1)^{2}

\frac{d}{d x} (\frac{4}{sin ( x ) + cos ( x )})

\int cos (x) - 9 sin (x) d x

\int \frac{5}{x + 5 + x} d x

What is the tangent of f(x)=5x^3,\at x=2 ?
The tangent of f(x)=5x^3,\at x=2 is y=60x-80