The value of [tex]\frac{dy}{dx}[/tex] is -3. [tex]\blacksquare[/tex]
We derive an expression for [tex]\frac{dy}{dx}[/tex] by means of chain rule and differentiation rule for a product of functions:
[tex]\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} ][/tex]
[tex]\frac{d}{dx}[f(x)\cdot g(y)] = f'(x)\cdot g(y) +f(x)\cdot g'(y)\cdot \frac{dy}{dx}[/tex] (1)
If we know that [tex]f(x) \cdot g(y) = 17-x-y[/tex], [tex]f(-2) = 3[/tex], [tex]f'(-2) = 4[/tex], [tex]g(4) = 5[/tex] and[tex]g'(4) = 2[/tex], then we have the following expression:
[tex]-1-\frac{dy}{dx} = (4)\cdot (5) + (3)\cdot (2) \cdot \frac{dy}{dx}[/tex]
[tex]-1-\frac{dy}{dx} = 20 + 6\cdot \frac{dy}{dx}[/tex]
[tex]7\cdot \frac{dy}{dx} = -21[/tex]
[tex]\frac{dy}{dx} = -3[/tex]
The value of [tex]\frac{dy}{dx}[/tex] is -3. [tex]\blacksquare[/tex]
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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 [tex]f(x)\cdot g(y) = 17 - x\cdot y[/tex], where [tex]f[/tex] is a differentiable function of [tex]x[/tex] and [tex]g[/tex] is a differentiable function of [tex]y[/tex]. Selected values of [tex]f[/tex], [tex]f'[/tex], [tex]g[/tex] and [tex]g'[/tex] are given below: [tex]f(-2) = 3[/tex], [tex]f'(-2) = 4[/tex], [tex]g(4) = 5[/tex], [tex]g'(4) = 2[/tex].
What is the value of [tex]\frac{dy}{dx}[/tex] at the point [tex](-2, 4)[/tex]?
