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

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

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tangent of f(x)=tan(x)

tangent of

f (x) = tan (x)

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

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

Compute the slope of

f (x) = tan (x)

x = a_{0} : m = sec^{2} (a_{0})

Find the line with slope m=

sec^{2} (a_{0})

and passing through

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

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

\int_{0}^{1} (3 x^{2} + 2 x + 1) d x

\frac{d}{d x} (\frac{2}{1 - x})

\int_{- \infty}^{0} \frac{1}{3 - 8 x} d x

\frac{\partial}{\partial x} (tan (x + y))

\int ln (r^{2}) d r

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