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jarptica [38.1K]
2 years ago
14

The point (−2,4) lies on the curve in the xy-plane given by the equation f(x)g(y)=17−x−y, where f is a differentiable function o

f x and g is a differentiable function of y. Selected values of f, f′, g, and g′ are given in the table above. What is the value of dydx at the point (−2,4) ?.
SAT
1 answer:
Nikitich [7]2 years ago
3 0

The value of \frac{dy}{dx} is -3. \blacksquare

<h2>Procedure - Differentiability</h2><h3 /><h3>Chain rule and derivatives</h3><h3 />

We derive an expression for \frac{dy}{dx} by means of chain rule and differentiation rule for a product of functions:

\frac{d}{dx}[f(x)\cdot g(y)] = [f'(x)\cdot \frac{dx}{dx}]\cdot g(y) + f(x) \cdot [g'(y)\cdot \frac{dy}{dx} ]

\frac{d}{dx}[f(x)\cdot g(y)] = f'(x)\cdot g(y) +f(x)\cdot g'(y)\cdot \frac{dy}{dx} (1)

If we know that f(x) \cdot g(y) = 17-x-y, f(-2) = 3,<em> </em>f'(-2) = 4,<em> </em>g(4) = 5 andg'(4) = 2, then we have the following expression:

-1-\frac{dy}{dx} = (4)\cdot (5) + (3)\cdot (2) \cdot \frac{dy}{dx}

-1-\frac{dy}{dx} = 20 + 6\cdot \frac{dy}{dx}

7\cdot \frac{dy}{dx} = -21

\frac{dy}{dx} = -3

The value of \frac{dy}{dx} is -3. \blacksquare

To learn more on differentiability, we kindly invite to check this verified question: brainly.com/question/24062595

<h3>Remark</h3>

The statement is incomplete and full of mistakes. Complete and corrected form is presented below:

<em>The point (-2, 4) lies on the curve in the xy-plane given by the equation </em>f(x)\cdot g(y) = 17 - x\cdot y<em>, where </em>f<em> is a differentiable function of </em>x<em> and </em>g<em> is a differentiable function of </em>y<em>. Selected values of </em>f<em>, </em>f'<em>, </em>g<em> and </em>g'<em> are given below: </em>f(-2) = 3<em>, </em>f'(-2) = 4<em>, </em>g(4) = 5<em>, </em>g'(4) = 2<em>. </em>

<em />

<em>What is the value of </em>\frac{dy}{dx}<em> at the point </em>(-2, 4)<em>?</em>

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