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

The tangent of f(x)=(1+2x)^2,\at x=4 is y=36x-63

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

tangent of

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

y = 36 x - 63

Solution steps

Find the tangent point:

(4, 81)

Find the slope of

f (x) = (1 + 2 x)^{2} : \frac{df ( x )}{d x} = 4 (1 + 2 x)

Compute the slope of

f (x) = (1 + 2 x)^{2}

x = 4 : m = 36

Find the line with slope m=

36

and passing through

(4, 81) : y = 36 x - 63

y = 36 x - 63

\frac{x ^{3}}{2 - x ^{2}}

(sin^{4} (x))^{^{'}}

\frac{2}{( s + 1 ) ^{3}}

\frac{d}{d x} (x e^{\frac{y}{x}})

x \to 1 lim ((1 - x)^{2} + 2)

What is the tangent of f(x)=(1+2x)^2,\at x=4 ?
The tangent of f(x)=(1+2x)^2,\at x=4 is y=36x-63