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

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

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

tangent of

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

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

Solution steps

Find the tangent point:

(2, 5^{\frac{1}{5}})

Find the slope of

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

Compute the slope of

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

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

Find the line with slope m=

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

and passing through

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

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

n = 0 \sum \infty \frac{n}{( n ^{2} + 3 )}

\frac{d}{d y} (2 x y)

f (x) = - \frac{12}{s ^{5}}

\frac{4}{3} π \cdot r^{2}

y^{^{''}} - 4 y^{^{'}} + 5 y = 36 cos (3 t) - 28 sin (3 t)

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