What is the tangent of f(x)=x*cot(x),\at x= pi/4 ?

The tangent of f(x)=x*cot(x),\at x= pi/4 is y=(1-pi/2)x+(pi^2)/8

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tangent of f(x)=x*cot(x),\at x= pi/4

tangent of

f (x) = x \cdot cot (x), at x = \frac{π}{4}

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

Solution steps

Find the tangent point:

(\frac{π}{4}, \frac{π}{4})

Find the slope of

f (x) = x cot (x) : \frac{df ( x )}{d x} = cot (x) - x csc^{2} (x)

Compute the slope of

f (x) = x cot (x)

x = \frac{π}{4} : m = 1 - \frac{π}{2}

Find the line with slope m=

1 - \frac{π}{2}

and passing through

(\frac{π}{4}, \frac{π}{4}) : y = (1 - \frac{π}{2}) x + \frac{π ^{2}}{8}

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

f (x) = \frac{x + x}{x ^{3}}

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

\frac{d}{d x} (e^{x^{2}} sin (x))

\frac{d x}{d t} = 2 x (10 - x)

\frac{\partial}{\partial t} (e^{t} sin (t))

What is the tangent of f(x)=x*cot(x),\at x= pi/4 ?
The tangent of f(x)=x*cot(x),\at x= pi/4 is y=(1-pi/2)x+(pi^2)/8