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

Find the tangent at

x = 1

Let's differentiate:

\frac{d}{d x} (8 \sqrt[4]{x})

ⓘ A constant factor stays in front of the derivative, (k*u)' = k*u':

\frac{d}{d x} (8 \sqrt[4]{x})

8 \cdot \frac{d}{d x} (\sqrt[4]{x})

■The derivative of x^n is n*x^(n-1):

\frac{d}{d x} (\sqrt[4]{x})

\frac{1}{4} \cdot x^{- \frac{3}{4}}

8 \cdot \frac{1}{4} \cdot x^{- \frac{3}{4}}

ⅰ

8 \cdot \frac{1}{4}

2

2 x^{- \frac{3}{4}}

ⓘ The factor is moved to the numerator or the denominator by changing the sign of its exponent, for a nonzero base:

2 x^{- \frac{3}{4}}

\frac{2}{\sqrt[4]{x^{3}}}

\frac{2}{\sqrt[4]{x^{3}}}

Compute f(

1

)

8 \cdot \sqrt[4]{1}

8

The point of tangency is (

1

8

)

Compute f'(

1

)

\frac{2}{\sqrt[4]{1^{3}}}

2

The slope is f'(

1

) =

2

The tangent passes through this point with this slope:

y = 2 \cdot (x - 1) + 8

Compute the y-intercept:

b = 8 - 2

8 - 2

ⓘ Start from

8

and move back

2

on the number line:

8 - 2

6

6

▶Tangent line:

y = 2 x + 6