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Calculus BC Implicit Differentiation Section 2.5 #9

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Calculus BC Implicit Differentiation Section 2.5 #9
by selina z - Friday, 1 July 2022, 12:47 PM


While rereading the textbook, I was sort of confused by the use of d/dx. For example in Section 2.5 #9, I used example 5 as a sample so I did d/dx only, so when I differentiated y with d/dx, I wound up with 1 * dy/dx. Would you mind clarifying the implicit differentiation process? Thank you!

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Re: Calculus BC Implicit Differentiation Section 2.5 #9
by Dr. Kevin Wang - Friday, 1 July 2022, 2:00 PM

From what you wrote, if you differentiate $y$ with $d/dx$, you get $dy/dx$, which is correct.

For implicit differentiation, just apply $d/dx$ to both sides of the equation, while keeping in mind that $y$ is a function of $x$, and apply the appropriate rules (product rule, chain rule, etc.).  For example, if the equation is $x^3 - xy + y^2 = 7$, you differentiate both sides to get

$$\frac{d}{dx}(x^3) - \frac{d}{dx}(xy) + \frac{d}{dx}(y^2) = 0.$$

The right hand side becomes $0$ because the original right hand side is $7$, a constant.  Now for the left hand side, the first term is simply the derivative of a power function.  The second term you apply the product rule: 

$$\frac{d}{dx}(xy) = \frac{dx}{dx} \cdot y + x\cdot \frac{dy}{dx} = y + x\cdot\frac{dy}{dx};$$

For the third term, you apply the chain rule:

$$\frac{d}{dx}(y^2) = 2y\cdot \frac{dy}{dx}.$$

Afterwards you get an equation about $\dfrac{dy}{dx}$, and get the answer by solving for it.