What is the tangent of y=sqrt(25-x^2),(3,4) ?

The tangent of y=sqrt(25-x^2),(3,4) is y=-3/4 x+25/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of y=sqrt(25-x^2),(3,4)

tangent of

y = 25 - x^{2}, (3, 4)

y = - \frac{3}{4} x + \frac{25}{4}

Solution steps

Find the slope of

y = 25 - x^{2} : \frac{d y}{d x} = - \frac{x}{25 - x ^{2}}

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{3}{4}

Find the line with slope m=

- \frac{3}{4}

and passing through

(3, 4) : y = - \frac{3}{4} x + \frac{25}{4}

y = - \frac{3}{4} x + \frac{25}{4}

\int \frac{e ^{- 2 x}}{e ^{- 2 x} + 1} d x

x \to - 2 lim (- x^{2} + 2 x - 6)

\int \frac{1}{x ^{3}} sin (\frac{1}{x ^{2}}) d x

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

\frac{d y}{d x} x^{2} y = y - 7

What is the tangent of y=sqrt(25-x^2),(3,4) ?
The tangent of y=sqrt(25-x^2),(3,4) is y=-3/4 x+25/4