What is the tangent of f(x)=tan(x),\at x=1 ?

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

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tangent of f(x)=tan(x),\at x=1

tangent of

f (x) = tan (x), at x = 1

y = sec^{2} (1) x + tan (1) - sec^{2} (1)

Solution steps

Find the tangent point:

(1, tan (1))

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 = 1 : m = sec^{2} (1)

Find the line with slope m=

sec^{2} (1)

and passing through

(1, tan (1)) : y = sec^{2} (1) x + tan (1) - sec^{2} (1)

y = sec^{2} (1) x + tan (1) - sec^{2} (1)

19 x, x^{2}

n = 0 \sum \infty \frac{1}{3} (\frac{2}{3})^{n}

y = lo g_{10} (x)

x \to 5 - lim (\frac{∣ 5 - x ∣}{5 - x})

\frac{( e ^{x} - 1 )}{x}

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