What is the tangent of f(x)=tan^2(x),\at x= pi/4 ?

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

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tangent of f(x)=tan^2(x),\at x= pi/4

tangent of

f (x) = tan^{2} (x), at x = \frac{π}{4}

y = 4 x + 1 - π

Solution steps

Find the tangent point:

(\frac{π}{4}, 1)

Find the slope of

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

Compute the slope of

f (x) = tan^{2} (x)

x = \frac{π}{4} : 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{sin ( x )}{3 x}

- lo g_{e} (1 - x)

y = 1 cos (2) (x)

e^{t^{2}}

y^{^{''}} + 2 y^{^{'}} + y = 18 e^{- t}

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