What is the tangent of f(x)=10(1+3x)^9,\at x=(0) ?

The tangent of f(x)=10(1+3x)^9,\at x=(0) is y=270x+10

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tangent of f(x)=10(1+3x)^9,\at x=(0)

tangent of

f (x) = 10 (1 + 3 x)^{9}, at x = (0)

y = 270 x + 10

Solution steps

Find the tangent point:

(0, 10)

Find the slope of

f (x) = 10 (1 + 3 x)^{9} : \frac{df ( x )}{d x} = 270 (1 + 3 x)^{8}

Compute the slope of

f (x) = 10 (1 + 3 x)^{9}

x = 0 : m = 270

Find the line with slope m=

270

and passing through

(0, 10) : y = 270 x + 10

y = 270 x + 10

f (x) = e^{x}, g (x) = 2 - e^{x}, [0, 3]

\frac{d}{d x} (x^{\frac{2}{3}} (x^{2} - 4))

n = 1 \sum \infty \frac{8 ^{n}}{9 ^{n}}

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

\frac{d}{d x} (1 - 2 sin (x))

What is the tangent of f(x)=10(1+3x)^9,\at x=(0) ?
The tangent of f(x)=10(1+3x)^9,\at x=(0) is y=270x+10