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) ?
Correct. The chain rule is the basis for implicit differentiation. f′(x)g(y)+f(x)g′(y)dydx=−1−dydxf′(x)g(y)+f(x)g′(y)dydx=−1−dydx (1+f(x)g′(y))dydx=−1−f′(x)g(y)⇒dydx=−1−f′(x)g(y)1+f(x)g′(y)(1+f(x)g′(y))dydx=−1−f′(x)g(y)⇒dydx=−1−f′(x)g(y)1+f(x)g′(y) dydx∣∣(−2,4)=−1−f′(−2)g(4)1+f(−2)g′(4)=−1−201+6=−217=−3
Let’s start by applying implicit differentiation to the given equation:
f(x)g(y) = 17 – x – y
Taking the derivative with respect to x on both sides, we get:
f'(x)g(y) = -1
Taking the derivative with respect to y on both sides, we get:
f(x)g'(y) = -1
Now, we need to find the values of f(-2), g(4), f'(-2), and g'(4) in order to evaluate dy/dx at the point (-2,4). From the given table:
f(-2) = 3, g(4) = 2, f'(-2) = -1/2, g'(4) = -1/2
Substituting these values into the equations we derived earlier, we get:
f'(-2)g(4) = -1/2 * 2 = -1
f(-2)g'(4) = 3 * (-1/2) = -3/2
Now we can use the chain rule to find dy/dx at the point (-2,4):
dy/dx = -g'(4)/f'(-2)
Plugging in the values we found earlier:
dy/dx = -(-1/2)/( -1/2)= 1
Therefore, the value of dy/dx at the point (-2,4) is 1.
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