The value of \frac{dy}{dx} is -3. \blacksquare
Procedure - DifferentiabilityChain 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 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
RemarkThe 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)?
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