What is the tangent of 4(8x-1)^4 ?

The tangent of 4(8x-1)^4 is y=128(8a_{0}-1)^3x-49152a_{0}^4+16384a_{0}^3-1536a_{0}^2+4

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tangent of 4(8x-1)^4

tangent of

4 (8 x - 1)^{4}

y = 128 (8 a_{0} - 1)^{3} x - 49152 a_{0}^{4} + 16384 a_{0}^{3} - 1536 a_{0}^{2} + 4

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 4 (8 a_{0} - 1)^{4})

Find the slope of

y = 4 (8 x - 1)^{4} : \frac{d y}{d x} = 128 (8 x - 1)^{3}

Compute the slope of

y = 4 (8 x - 1)^{4}

x = a_{0} : m = 128 (8 a_{0} - 1)^{3}

Find the line with slope m=

128 (8 a_{0} - 1)^{3}

and passing through

(a_{0}, 4 (8 a_{0} - 1)^{4}) : y = 128 (8 a_{0} - 1)^{3} x - 49152 a_{0}^{4} + 16384 a_{0}^{3} - 1536 a_{0}^{2} + 4

y = 128 (8 a_{0} - 1)^{3} x - 49152 a_{0}^{4} + 16384 a_{0}^{3} - 1536 a_{0}^{2} + 4

x^{^{''}} + 4 x^{^{'}} + 4 x = e^{- 2 t}

f (x) = 3 x^{3}

\int_{0}^{8} - x^{2} + 2 x + 4 d x

f (x) = \frac{4 x}{x + 1}, at x = 3

x \to - 4 lim (\frac{∣ x + 4 ∣}{x + 4})

What is the tangent of 4(8x-1)^4 ?
The tangent of 4(8x-1)^4 is y=128(8a_{0}-1)^3x-49152a_{0}^4+16384a_{0}^3-1536a_{0}^2+4