Answer this question and show the steps. x+y=3, find the value of dydx at the point (4,1)

Step 1 Consider the given function x+y=3 Step 2 Differentiate implicitly with respect to x. ddx(x+y)=ddx(3) ddx(x12)+ddx(y12)=0 12x12−1+(12y12−1)dydx=0 12x−12+(12y−12)dydx=0 12x+12ydydx=0 dydx=−yx=−yx Step 3 Now, find the value of dydx at the point (4, 1) dydx(x,y)=(4,1)=−14=−12 The answer is −12 or -0.5(in decimals).

"Answer this question and show the steps. x+y=3,..." was answered by students like you

melodykap

Answered question

2020-12-24

Answer this question and show the steps.

\sqrt{x} + \sqrt{y} = 3

, find the value of

\frac{d y}{d x}

at the point (4,1)

Answer & Explanation

crocolylec

Skilled2020-12-25Added 100 answers

Step 1 Consider the given function

\sqrt{x} + \sqrt{y} = 3

Step 2 Differentiate implicitly with respect to x.

\frac{d}{d x} (\sqrt{x} + \sqrt{y}) = \frac{d}{d x} (3)

\frac{d}{d x} (x^{\frac{1}{2}}) + \frac{d}{d x} (y^{\frac{1}{2}}) = 0

\frac{1}{2} x^{\frac{1}{2} - 1} + (\frac{1}{2} y^{\frac{1}{2} - 1}) \frac{d y}{d x} = 0

\frac{1}{2} x^{- \frac{1}{2}} + (\frac{1}{2} y^{- \frac{1}{2}}) \frac{d y}{d x} = 0

\frac{1}{2 \sqrt{x}} + \frac{1}{2 \sqrt{y}} \frac{d y}{d x} = 0

\frac{d y}{d x} = - \frac{\sqrt{y}}{\sqrt{x}} = - \sqrt{\frac{y}{x}}

Step 3 Now, find the value of

\frac{d y}{d x}

at the point (4, 1)

\frac{d y}{d x} (x, y) = (4, 1) = - \sqrt{\frac{1}{4}} = - \frac{1}{2}

The answer is

- \frac{1}{2}

or -0.5(in decimals).

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