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{d}{d x} (e^{\frac{1}{2} x^{2}})

\frac{d}{d x} (2^{3 x + 1} ln (5 x - 11))

e^{\frac{t}{2}} \cdot cos (2 t)

\int 9 sin (ln (x)) d x

\int 2 x e^{2 y} d y

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