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

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

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

tangent of

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

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

Solution steps

Find the tangent point:

(1, 6^{\frac{1}{4}})

Find the slope of

f (x) = (4 x + 2)^{\frac{1}{4}} : \frac{df ( x )}{d x} = \frac{1}{( 4 x + 2 ) ^{\frac{3}{4}}}

Compute the slope of

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

x = 1 : m = \frac{1}{6 ^{\frac{3}{4}}}

Find the line with slope m=

\frac{1}{6 ^{\frac{3}{4}}}

and passing through

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

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

\frac{\partial}{\partial x} (\frac{4}{4 x + y})

n = 1 \sum \infty \frac{2 n}{3}

\frac{\partial}{\partial x} (3 e^{y})

\frac{\partial}{\partial x} (sin (2 x) cos (2 t))

\int \frac{ln ( 2 x )}{ln ( 4 x ) x} d x

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