What is the tangent of f(x)=(2x+4)^{1/5},\at x=1 ?

The tangent of f(x)=(2x+4)^{1/5},\at x=1 is y=(2^{1/5})/(5*3^{4/5)}x+6^{1/5}-(2^{1/5})/(5*3^{4/5)}

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tangent of f(x)=(2x+4)^{1/5},\at x=1

tangent of

f (x) = (2 x + 4)^{\frac{1}{5}}, at x = 1

y = \frac{2 ^{\frac{1}{5}}}{5 \cdot 3 ^{\frac{4}{5}}} x + 6^{\frac{1}{5}} - \frac{2 ^{\frac{1}{5}}}{5 \cdot 3 ^{\frac{4}{5}}}

Solution steps

Find the tangent point:

(1, 6^{\frac{1}{5}})

Find the slope of

f (x) = (2 x + 4)^{\frac{1}{5}} : \frac{df ( x )}{d x} = \frac{2}{5 ( 2 x + 4 ) ^{\frac{4}{5}}}

Compute the slope of

f (x) = (2 x + 4)^{\frac{1}{5}}

x = 1 : m = \frac{2 ^{\frac{1}{5}}}{5 \cdot 3 ^{\frac{4}{5}}}

Find the line with slope m=

\frac{2 ^{\frac{1}{5}}}{5 \cdot 3 ^{\frac{4}{5}}}

and passing through

(1, 6^{\frac{1}{5}}) : y = \frac{2 ^{\frac{1}{5}}}{5 \cdot 3 ^{\frac{4}{5}}} x + 6^{\frac{1}{5}} - \frac{2 ^{\frac{1}{5}}}{5 \cdot 3 ^{\frac{4}{5}}}

y = \frac{2 ^{\frac{1}{5}}}{5 \cdot 3 ^{\frac{4}{5}}} x + 6^{\frac{1}{5}} - \frac{2 ^{\frac{1}{5}}}{5 \cdot 3 ^{\frac{4}{5}}}

y \to \infty lim (\frac{( y ^{2} - e ^{(6 y)} )}{( 4 y ^{2} + e ^{7 y} )})

h (t) = (t + 1)^{\frac{2}{3}} (2 t^{2} - 1)^{3}

\frac{\partial}{\partial y} (4 x^{3} y^{3} + 1)

n = 0 \sum \infty 0. 1^{n}

\int \frac{1}{5 x + 6} d x

What is the tangent of f(x)=(2x+4)^{1/5},\at x=1 ?
The tangent of f(x)=(2x+4)^{1/5},\at x=1 is y=(2^{1/5})/(5*3^{4/5)}x+6^{1/5}-(2^{1/5})/(5*3^{4/5)}