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

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

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tangent of f(x)=-sin(x),\at x=0

tangent of

f (x) = - sin (x), at x = 0

y = - x

Solution steps

Find the tangent point:

(0, 0)

Find the slope of

f (x) = - sin (x) : \frac{df ( x )}{d x} = - cos (x)

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 = - 1

Find the line with slope m=

- 1

and passing through

(0, 0) : y = - x

y = - x

e^{x + y} y^{^{'}} = x

x \to 5 + lim (ln (sin (x)))

y^{^{'}} = \frac{( x y + 2 y )}{x + 3}

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

\frac{\partial}{\partial x} (ln (1 + 3 x^{2} y^{3}))

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