What is the tangent of f(x)=x^2-tan(x),\at x=0 ?

The tangent of f(x)=x^2-tan(x),\at x=0 is y=-x

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tangent of f(x)=x^2-tan(x),\at x=0

tangent of

f (x) = x^{2} - tan (x), at x = 0

y = - x

Solution steps

Find the tangent point:

(0, 0)

Find the slope of

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

Compute the slope of

f (x) = x^{2} - tan (x)

x = 0 : m = - 1

Find the line with slope m=

- 1

and passing through

(0, 0) : y = - x

y = - x

\frac{\partial}{\partial y} (9 e^{yz} cos (x yz))

\int \frac{3 ln ^{2} ( x ) + 25}{x} d x

\int \frac{2 x}{y ^{2}} d x

sin (x cos (x))

\frac{d}{d x} (x^{2} e^{6 x})

What is the tangent of f(x)=x^2-tan(x),\at x=0 ?
The tangent of f(x)=x^2-tan(x),\at x=0 is y=-x