tangent of f(x)=3tan(x),\at x= pi/3

⚠Ⓘ 3tan is read as the product 3*tan(...): for a power, write 3^tan(...)

1) Find the tangent at

x = \frac{π}{3}

⇥1.1) Let's differentiate:

\frac{d}{d x} (3 \cdot \tan (x))

⇥ℹ

\frac{d}{d x} (3 \cdot \tan (x))

3 \cdot \frac{d}{d x} (\tan (x))

‖

\frac{d}{d x} (\tan (x))

\frac{1}{{\cos (x)}^{2}}

⇥

3 \cdot \frac{1}{{\cos (x)}^{2}}

⇥

\frac{3}{{\cos (x)}^{2}}

Compute f(

\frac{π}{3}

)

⇥

3 \cdot \tan (\frac{π}{3})

⇥

3 \cdot \sqrt{3}

The point of tangency is (

\frac{π}{3}

3 \cdot \sqrt{3}

)

Compute f'(

\frac{π}{3}

)

⇥

\frac{3}{{\cos (\frac{π}{3})}^{2}}

⇥

\frac{3}{{(\frac{1}{2})}^{2}}

⇥

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

⇥

\frac{3}{\frac{1}{4}}

⇥

3 \cdot 4

⇥

12

The slope is f'(

\frac{π}{3}

) =

12

The tangent passes through this point with this slope:

y = 12 \cdot (x - \frac{π}{3}) + 3 \cdot \sqrt{3}

Compute the y-intercept:

b = 3 \cdot \sqrt{3} - 12 \cdot \frac{π}{3}

⇥

3 \cdot \sqrt{3} - 12 \cdot \frac{π}{3}

⇥ℹ

- 12 \cdot \frac{π}{3}

- \frac{12 π}{3}

‖

- \frac{12 π}{3}

- 4 π

⇥

3 \cdot \sqrt{3} - 4 π

▶Tangent line:

y = 12 x + 3 \cdot \sqrt{3} - 4 π