What is the tangent of y=10cot((pix)/(20)),\at x=5 ?

The tangent of y=10cot((pix)/(20)),\at x=5 is y=-pix+10+5pi

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tangent of y=10cot((pix)/(20)),\at x=5

tangent of

y = 10 cot (\frac{π x}{20}), at x = 5

y = - π x + 10 + 5 π

Solution steps

Find the tangent point:

(5, 10)

Find the slope of

y = 10 cot (\frac{π x}{20}) : \frac{d y}{d x} = - \frac{π}{2} csc^{2} (\frac{π x}{20})

Compute the slope of

y = 10 cot (\frac{π x}{20})

x = 5 : m = - π

Find the line with slope m=

- π

and passing through

(5, 10) : y = - π x + 10 + 5 π

y = - π x + 10 + 5 π

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2 \frac{d ^{2} y}{d x ^{2}} + 8 y = cos (2 x)

\frac{\partial}{\partial x} (2 (x - 1))

\int \frac{1}{( x + 3 ) ^{3}} d x

What is the tangent of y=10cot((pix)/(20)),\at x=5 ?
The tangent of y=10cot((pix)/(20)),\at x=5 is y=-pix+10+5pi