Question

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 of 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) ?.

Answer 1

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

Procedure - Differentiability

Chain rule and derivatives

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 $f(x) \cdot g(y) = 17-x-y$, $f(-2) = 3$, $f'(-2) = 4$, $g(4) = 5$ and$g'(4) = 2$, 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. $\blacksquare$

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Remark

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

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

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