What is the tangent of f(x)=8\sqrt[4]{x},\at x=1 ?

The tangent of f(x)=8\sqrt[4]{x},\at x=1 is y=2x+6

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tangent of f(x)=8\sqrt[4]{x},\at x=1

tangent of

y = 2 x + 6

Solution steps

Find the tangent point:

(1, 8)

Find the slope of

Compute the slope of

x = 1 : m = 2

Find the line with slope m=

2

and passing through

(1, 8) : y = 2 x + 6

y = 2 x + 6

\frac{d}{d x} ((3 x + 5)^{3} - 5 x)

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

\int \frac{1}{cos ^{2} ( b ) x} d x

x \to 2 lim (- 5)

\int - cos (2 t) d t

What is the tangent of f(x)=8\sqrt[4]{x},\at x=1 ?
The tangent of f(x)=8\sqrt[4]{x},\at x=1 is y=2x+6