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

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

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tangent of f(x)=(6x-2)^{1/4},\at x=1

tangent of

f (x) = (6 x - 2)^{\frac{1}{4}}, at x = 1

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

Solution steps

Find the tangent point:

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

Find the slope of

f (x) = (6 x - 2)^{\frac{1}{4}} : \frac{df ( x )}{d x} = \frac{3}{2 ( 6 x - 2 ) ^{\frac{3}{4}}}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = \frac{3}{4 \cdot 2 ^{\frac{1}{2}}}

Find the line with slope m=

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

and passing through

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

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

\frac{8}{t}

\int x sin (x) cos (x) d x

\frac{d}{d x} ((2 - x)^{2})

\frac{\partial}{\partial x} (4 x y - 10 x^{2} y^{2} z + 5 z^{2})

y^{\frac{4}{5}}

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