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)

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : 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

f (x) = 13 x e^{x}

y = 3 1 + 8 x

\int \frac{1}{2 π} e^{- \frac{x ^{2}}{2}} d x

\int \frac{5 x + 12}{x ( x ^{2} + 4 )} d x

\int (x + 1)^{- 8} x d x

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