What is the normal of y=(sqrt(x))/(x+4),(1,0.2) ?

The normal of y=(sqrt(x))/(x+4),(1,0.2) is y=-50/3 x+16.86666…

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normal of y=(sqrt(x))/(x+4),(1,0.2)

normal of

y = \frac{x}{x + 4}, (1, 0.2)

y = - \frac{50}{3} x + 16.86666 \dots

Solution steps

Find the slope of

y = \frac{x}{x + 4} : \frac{d y}{d x} = \frac{- x + 4}{2 x ( x + 4 ) ^{2}}

Compute the slope of

y = \frac{x}{x + 4}

(1, 0.2) : m = \frac{3}{50}

Compute the slope of the perpendicular line:

m_{p} = - \frac{50}{3}

Find the line with slope m=

- \frac{50}{3}

and passing through

(1, 0.2) : y = - \frac{50}{3} x + 16.86666 \dots

y = - \frac{50}{3} x + 16.86666 \dots

y = \frac{1}{x}, y = x, y = \frac{1}{4} x

\int x \cdot arcsin (x) d x

(x^{2} + x^{3})^{4}

\int \frac{t}{( t ^{2} + 1 ) ^{\frac{7}{2}}} d t

y = 2 x^{2} - 1, (0, 3)

What is the normal of y=(sqrt(x))/(x+4),(1,0.2) ?
The normal of y=(sqrt(x))/(x+4),(1,0.2) is y=-50/3 x+16.86666…