What is the tangent of y=tan^2(x),(-pi/4 ,1) ?

The tangent of y=tan^2(x),(-pi/4 ,1) is y=-4x+1-pi

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tangent of y=tan^2(x),(-pi/4 ,1)

tangent of

y = tan^{2} (x), (- \frac{π}{4}, 1)

y = - 4 x + 1 - π

Solution steps

Find the slope of

y = tan^{2} (x) : \frac{d y}{d x} = 2 tan (x) sec^{2} (x)

Compute the slope of

y = tan^{2} (x)

(- \frac{π}{4}, 1) : m = - 4

Find the line with slope m=

- 4

and passing through

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

y = - 4 x + 1 - π

\frac{\partial}{\partial y} (\frac{8 x}{y})

\frac{s}{s ^{2} - 2 s + 4}

\frac{\partial}{\partial u} (3 sin (v) cos (u))

\frac{d}{d x} (\frac{3}{4 + x ^{2}})

\frac{( s - 2 )}{s ^{2} ( s - 1 ) ^{2}}

What is the tangent of y=tan^2(x),(-pi/4 ,1) ?
The tangent of y=tan^2(x),(-pi/4 ,1) is y=-4x+1-pi