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

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

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

tangent of

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

y = 45 x + 1

Solution steps

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}

(0, 1) : m = 45

Find the line with slope m=

45

and passing through

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

y = 45 x + 1

e^{- x}, 1

\frac{\partial}{\partial x} ((cos (x))^{2})

- \frac{1}{( x - 6 ) ^{2}}

\int csc (4 x) d x

\frac{d}{d x} (arcsinh (x))

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