What is the tangent of f(x)=(1+5x)^9,\at x=0 ?

The tangent of f(x)=(1+5x)^9,\at x=0 is y=45x+1

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tangent of f(x)=(1+5x)^9,\at x=0

tangent of

f (x) = (1 + 5 x)^{9}, at x = 0

y = 45 x + 1

Solution steps

Find the tangent point:

(0, 1)

Find the slope of

f (x) = (1 + 5 x)^{9} : \frac{df ( x )}{d x} = 45 (1 + 5 x)^{8}

Compute the slope of

f (x) = (1 + 5 x)^{9}

x = 0 : m = 45

Find the line with slope m=

45

and passing through

(0, 1) : y = 45 x + 1

y = 45 x + 1

x \to 0 + lim (\frac{I n x}{x})

\frac{d y}{d x} = \frac{( x y + 3 x - y - 3 )}{x y - 2 x + 4 y - 8}

\int \frac{2 x}{x ^{2} + x} d x

\frac{s}{s + 4}

\frac{d y}{d x} = \frac{1}{x + y}

What is the tangent of f(x)=(1+5x)^9,\at x=0 ?
The tangent of f(x)=(1+5x)^9,\at x=0 is y=45x+1