What is the tangent of f(x)=tan(x),\at x=0 ?

The tangent of f(x)=tan(x),\at x=0 is y=x

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tangent of f(x)=tan(x),\at x=0

tangent of

f (x) = tan (x), at x = 0

y = x

Solution steps

Find the tangent point:

(0, 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 = 0 : m = 1

Find the line with slope m=

1

and passing through

(0, 0) : y = x

y = x

y^{^{'''}} + y^{^{''}} - 8 y^{^{'}} - 12 y = 0

(t (x) - t (x)^{3}) v^{'} + 2 v = 0

\frac{\partial}{\partial x} (x y e^{x - y})

\frac{7}{x ^{6}}

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

What is the tangent of f(x)=tan(x),\at x=0 ?
The tangent of f(x)=tan(x),\at x=0 is y=x