What is the tangent of f(x)=tan(x+4) ?

The tangent of f(x)=tan(x+4) is y=sec^2(a_{0}+4)x+tan(a_{0}+4)-sec^2(a_{0}+4)a_{0}

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tangent of f(x)=tan(x+4)

tangent of

f (x) = tan (x + 4)

y = sec^{2} (a_{0} + 4) x + tan (a_{0} + 4) - sec^{2} (a_{0} + 4) a_{0}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, tan (a_{0} + 4))

Find the slope of

f (x) = tan (x + 4) : \frac{df ( x )}{d x} = sec^{2} (x + 4)

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 = sec^{2} (a_{0} + 4)

Find the line with slope m=

sec^{2} (a_{0} + 4)

and passing through

(a_{0}, tan (a_{0} + 4)) : y = sec^{2} (a_{0} + 4) x + tan (a_{0} + 4) - sec^{2} (a_{0} + 4) a_{0}

y = sec^{2} (a_{0} + 4) x + tan (a_{0} + 4) - sec^{2} (a_{0} + 4) a_{0}

\frac{3 - x e ^{x}}{x + e ^{x}}

\frac{d}{d x} (x cos (ln (x)))

\int 2 cos (θ) d θ

5 e^{x} cos (x)

\frac{d}{d x} (- sin (\frac{2}{x + 1} - e^{x^{2}} \cdot 2 x))

What is the tangent of f(x)=tan(x+4) ?
The tangent of f(x)=tan(x+4) is y=sec^2(a_{0}+4)x+tan(a_{0}+4)-sec^2(a_{0}+4)a_{0}