Question 7.1.2: The graph of y³ + 3x²y = 13 (∗) was studied in Example 5.4.2......
The graph of
y³ + 3x²y = 13 (∗)
was studied in Example 5.4.2 and is drawn larger in Fig. 7.1.2. It passes through the point (2, 1). Find the slope of the graph at that point.
Since in this case there is no simple way of expressing y as an explicit function of x, we use implicit differentiation.We think of replacing y with an unspecified function of x wherever y occurs. Then y³ + 3x²y becomes a function of x which is equal to the constant 13 for all x. So the derivative of y³ + 3x²y w.r.t. x must be equal to zero for all x. According to the chain rule, the derivative of y³ w.r.t. x is equal to 3y^{2}y^{\prime}. Using the product rule, the
derivative of 3x²y is equal to 6xy + 3x^{2}y^{\prime}. Hence, differentiating (∗) gives
F(x)=f(x)\cdot g(x)\Rightarrow F^{\prime}(x)=f^{\prime}(x)\cdot g(x)+f(x)\cdot g^{\prime}(x) the product rule
3y^{2}y^{\prime}+6x y+3x^{2}y^{\prime}=0 (∗∗)
Solving this equation for y^{\prime} yields
y^{\prime}={\frac{-6x y}{3x^{2}+3y^{2}}}={\frac{-2x y}{x^{2}+y^{2}}} (7.1.1)
For x = 2, y = 1 we find y^{\prime} = −4/5, which agrees with Fig. 7.1.2.^{1}
¹ Recall Fig. 5.4.3.
