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

The tangent of f(x)=5x^2,\at x=2 is y=20x-20

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

tangent of

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

y = 20 x - 20

Solution steps

Find the tangent point:

(2, 20)

Find the slope of

f (x) = 5 x^{2} : \frac{df ( x )}{d x} = 10 x

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 = 20

Find the line with slope m=

20

and passing through

(2, 20) : y = 20 x - 20

y = 20 x - 20

\frac{d}{d x} (\frac{1 + sin ( x )}{x cos ( x )})

x \to 1 lim (3^{(\frac{1}{x})})

3 sin (3 t + 2)

\int 4 x^{3} e^{2 x^{4}} d x

\frac{d}{d x} ((\frac{x ^{2}}{6 - x})^{5})

What is the tangent of f(x)=5x^2,\at x=2 ?
The tangent of f(x)=5x^2,\at x=2 is y=20x-20