What is the equation of the line tangent to f(x)=xtan⁡2x at x=π8?

Explanation:Product Rule:f′(x)=(1)(tan⁡(2x))+(x)(ddx[tan⁡(2x)])f′(x)=tan⁡(2x)+(x)[(2sec2(2x))]f′(x)=tan⁡(2x)+2xsec2(x)Evaluate the slope when x=π8f′(π8)=tan⁡(2(π8))+2(π8)sec2(2(π8))m=1+π2Evaluate the function when x=π8f(x)=xtan⁡(2x)=(π8)tan⁡(2(π8))=π8Putting it all together:y−y1=m(x−x1)y−π8=(π2+1)(x−π8)y=(π2+1)(x−π8)y=(π2+1)(x−π8)+π8

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High School
Calculus and Analysis
Calculus 1
Derivatives

Haylee Bowen

Answered question

2022-04-07

What is the equation of the line tangent to

f (x) = x \tan 2 x

x = \frac{π}{8}

Answer & Explanation

rhyclelal80j6

Beginner2022-04-08Added 13 answers

Explanation:
Product Rule:

f^{'} (x) = (1) (\tan (2 x)) + (x) (\frac{d}{d x} [\tan (2 x)])

f^{'} (x) = \tan (2 x) + (x) [(2 \sec^{2} (2 x))]

f^{'} (x) = \tan (2 x) + 2 x \sec^{2} (x)

Evaluate the slope when

x = \frac{π}{8}

f^{'} (\frac{π}{8}) = \tan (2 (\frac{π}{8})) + 2 (\frac{π}{8}) \sec^{2} (2 (\frac{π}{8}))

m = 1 + \frac{π}{2}

Evaluate the function when

x = \frac{π}{8}

f (x) = x \tan (2 x) = (\frac{π}{8}) \tan (2 (\frac{π}{8})) = \frac{π}{8}

Putting it all together:

y - y_{1} = m (x - x_{1})

y - \frac{π}{8} = (\frac{π}{2} + 1) (x - \frac{π}{8})

y = (\frac{π}{2} + 1) (x - \frac{π}{8})

y = (\frac{π}{2} + 1) (x - \frac{π}{8}) + \frac{π}{8}

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