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

The tangent of y=f(x),\at x=-1 is y= d/(d(-1))(f((-1)))x+f((-1))+d/(d(-1))(f((-1)))

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tangent of y=f(x),\at x=-1

tangent of

y = f (x), at x = - 1

y = \frac{d}{d ( - 1 )} (f ((- 1))) x + f ((- 1)) + \frac{d}{d ( - 1 )} (f ((- 1)))

Solution steps

Find the tangent point:

(- 1, f ((- 1)))

Find the slope of

y = f (x) : \frac{d y}{d x} = \frac{df}{d x}

Compute the slope of

y = f (x)

x = - 1 : m = \frac{d}{d ( - 1 )} (f ((- 1)))

Find the line with slope m=

\frac{d}{d ( - 1 )} (f ((- 1)))

and passing through

(- 1, f ((- 1))) : y = \frac{d}{d ( - 1 )} (f ((- 1))) x + f ((- 1)) + \frac{d}{d ( - 1 )} (f ((- 1)))

y = \frac{d}{d ( - 1 )} (f ((- 1))) x + f ((- 1)) + \frac{d}{d ( - 1 )} (f ((- 1)))

x \to 2 lim (x^{2})

\int - x^{2} e^{x} d x

y = 6 - 6 x^{2}, y = 0

f (x) = \frac{2 x}{ln ( x )}

x \to 0 lim ((cos (x))^{c s c (x)})

What is the tangent of y=f(x),\at x=-1 ?
The tangent of y=f(x),\at x=-1 is y= d/(d(-1))(f((-1)))x+f((-1))+d/(d(-1))(f((-1)))