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

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

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tangent of y=(1+3x)^{10},(0,1)

tangent of

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

y = 30 x + 1

Solution steps

Find the slope of

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

Compute the slope of

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

(0, 1) : m = 30

Find the line with slope m=

30

and passing through

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

y = 30 x + 1

x \to 0 lim (2 e^{5 x})

\int_{0}^{7} - 30 x^{2} + 340 x d x

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

(x^{2} - y^{2}) d x + 3 x y d y = 0

\int_{4}^{7} \frac{1}{x - 3} d x

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