What is the tangent of y=sqrt(9-10x),\at x=-1 ?

The tangent of y=sqrt(9-10x),\at x=-1 is y=-5/(sqrt(19))x+sqrt(19)-5/(sqrt(19))

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tangent of y=sqrt(9-10x),\at x=-1

tangent of

y = 9 - 10 x, at x = - 1

y = - \frac{5}{19} x + 19 - \frac{5}{19}

Solution steps

Find the tangent point:

(- 1, 19)

Find the slope of

y = 9 - 10 x : \frac{d y}{d x} = - \frac{5}{9 - 10 x}

Compute the slope of

y = 9 - 10 x

x = - 1 : m = - \frac{5}{19}

Find the line with slope m=

- \frac{5}{19}

and passing through

(- 1, 19) : y = - \frac{5}{19} x + 19 - \frac{5}{19}

y = - \frac{5}{19} x + 19 - \frac{5}{19}

\frac{d}{d x} (2 x y^{3})

\frac{\partial}{\partial y} (e^{s i n (x + y)})

f (x) = 1 + 4 csc (x), at x = - \frac{π}{3}

n = 1 \sum \infty \frac{2}{3 n + 10}

\int_{1}^{2} 4 x 4 x^{2} + 4 d x

What is the tangent of y=sqrt(9-10x),\at x=-1 ?
The tangent of y=sqrt(9-10x),\at x=-1 is y=-5/(sqrt(19))x+sqrt(19)-5/(sqrt(19))