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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) ?.

Sagot :

xero099

The value of [tex]\frac{dy}{dx}[/tex] is -3. [tex]\blacksquare[/tex]

Procedure - Differentiability

Chain rule and derivatives

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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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 [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]?

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