What is the tangent of f(x)=4x^2,\at x=1 ?

The tangent of f(x)=4x^2,\at x=1 is y=8x-4

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tangent of f(x)=4x^2,\at x=1

tangent of

f (x) = 4 x^{2}, at x = 1

y = 8 x - 4

Solution steps

Find the tangent point:

(1, 4)

Find the slope of

f (x) = 4 x^{2} : \frac{df ( x )}{d x} = 8 x

Compute the slope of

f (x) = 4 x^{2}

x = 1 : m = 8

Find the line with slope m=

8

and passing through

(1, 4) : y = 8 x - 4

y = 8 x - 4

\frac{\partial}{\partial x} (x^{9 y})

\frac{\partial}{\partial u} (u \cdot cos (v))

f (x) = arccos (x)

y = 2 sin (π x - y)

\int_{3}^{- 1} x^{3} d x

What is the tangent of f(x)=4x^2,\at x=1 ?
The tangent of f(x)=4x^2,\at x=1 is y=8x-4