What is the tangent of f(x)=sqrt(1-4x),\at x=-1 ?

The tangent of f(x)=sqrt(1-4x),\at x=-1 is y=-2/(sqrt(5))x+sqrt(5)-2/(sqrt(5))

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tangent of f(x)=sqrt(1-4x),\at x=-1

tangent of

f (x) = 1 - 4 x, at x = - 1

y = - \frac{2}{5} x + 5 - \frac{2}{5}

Solution steps

Find the tangent point:

(- 1, 5)

Find the slope of

f (x) = 1 - 4 x : \frac{df ( x )}{d x} = - \frac{2}{1 - 4 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 = - \frac{2}{5}

Find the line with slope m=

- \frac{2}{5}

and passing through

(- 1, 5) : y = - \frac{2}{5} x + 5 - \frac{2}{5}

y = - \frac{2}{5} x + 5 - \frac{2}{5}

(x + x) \frac{d y}{d x} = y + y

(2 x y^{2} + 7 y) + (2 x^{2} y + 7 x) y^{^{'}} = 0

x^{2} y^{^{''}} + 2 x y^{^{'}} - 2 = 0

\frac{d}{d x} (y^{2} - x^{2})

\frac{\partial}{\partial x} (x^{0.25} y^{0.5})

What is the tangent of f(x)=sqrt(1-4x),\at x=-1 ?
The tangent of f(x)=sqrt(1-4x),\at x=-1 is y=-2/(sqrt(5))x+sqrt(5)-2/(sqrt(5))