What is the tangent of y=(1+4x)^{10},(0,1) ?

The tangent of y=(1+4x)^{10},(0,1) is y=40x+1

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tangent of y=(1+4x)^{10},(0,1)

tangent of

y = (1 + 4 x)^{10}, (0, 1)

y = 40 x + 1

Solution steps

Find the slope of

y = (1 + 4 x)^{10} : \frac{d y}{d x} = 40 (1 + 4 x)^{9}

Compute the slope of

y = (1 + 4 x)^{10}

(0, 1) : m = 40

Find the line with slope m=

40

and passing through

(0, 1) : y = 40 x + 1

y = 40 x + 1

3 \frac{d y}{d x} = e^{y - 6 x}

\frac{d}{d x} (2 (2 x e^{2 x} + e^{2 x} \cdot 2 x^{2}))

x \to - 2 lim (\frac{x}{( x + 2 ) ^{3}})

\int x sec (x^{2}) tan (x^{2}) d x

f (x) = \frac{2 x ^{3} - 4 x ^{2} + 3}{x}

What is the tangent of y=(1+4x)^{10},(0,1) ?
The tangent of y=(1+4x)^{10},(0,1) is y=40x+1