What is the tangent of f(x)=(1+3x)^{10},\at x=0 ?

The tangent of f(x)=(1+3x)^{10},\at x=0 is y=30x+1

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tangent of f(x)=(1+3x)^{10},\at x=0

tangent of

f (x) = (1 + 3 x)^{10}, at x = 0

y = 30 x + 1

Solution steps

Find the tangent point:

(0, 1)

Find the slope of

f (x) = (1 + 3 x)^{10} : \frac{df ( x )}{d x} = 30 (1 + 3 x)^{9}

Compute the slope of

f (x) = (1 + 3 x)^{10}

x = 0 : m = 30

Find the line with slope m=

30

and passing through

(0, 1) : y = 30 x + 1

y = 30 x + 1

x \to 2 lim (x^{2} + 4 x)

2 y \frac{d y}{d x} + 2 = y^{2} + 2 x

\frac{d}{d x} ((2 x - 2)^{3} (3 x - 1)^{2})

x \to 0 + lim (\frac{1 - ln ( x )}{e ^{\frac{1}{x}}})

\frac{d}{d x} (2 x^{2} y - x^{3})

What is the tangent of f(x)=(1+3x)^{10},\at x=0 ?
The tangent of f(x)=(1+3x)^{10},\at x=0 is y=30x+1