# 数学代写|微积分代写Calculus代写|MATH141

## 数学代写|微积分代写Calculus代写|Differentiating implicitly defined functions

Although the curve $x^2+y^2=1$ is not a function, it is a circle and therefore does have a tangent line at any given point. It even looks as though we should be able to find the tangent. If we zoom in near the point of tangency, the picture looks like the graph of a function with a tangent line (figure 2).

Figure 2 illustrates the key idea: keep only a portion of the curve so that what remains passes the vertical line test and therefore represents a function. This can be done in many different ways. For instance, we could keep the top half of the curve or the bottom half of the curve, as in figure 3. Either way, what remains passes the vertical line test and
Figure 3 The graphs of two functions defined implicitly by $x^2+y^2=1$
represents the graph of a function. These are not the only two possibilities; we could also keep the top left and bottom right quarters of the circle, or even pick and choose other pieces seemingly at random, as long as what remains passes the vertical line test. See figure 4 . There is a sense in which the equation $x^2+y^2=1$ defines all of these different functions. These functions are not written explicitly, but they are implied; therefore, we say that these functions are defined implicitly by the equation $x^2+y^2=1$.

There are two ideas of how we might proceed to find the slope of the tangent line. One is to determine the equation of the bottom half of the circle and then use our previous methods (find the derivative and evaluate it) to find the slope. This is demonstrated later (example 6), but the problem with this method is that it is sometimes hard or even impossible to find a function representing the appropriate piece of the curve. We need a different method.

The second idea, called implicit differentiation, is to go ahead and take the derivative of the equation in its current form. We need rules for differentiating an equation such as $x^2+y^2=1$, instead of a function.
If we want the slope of the tangent line, we want to know $\frac{d y}{d x}$. That is, we want to find the derivative with respect to the variable $x$. The variables $x$ and $y$ are not treated the same:
$$\frac{d}{d x} x=1,$$
whereas
$$\frac{d}{d x} y=\frac{d y}{d x}=y^{\prime}$$

## 数学代写|微积分代写Calculus代写|Implicit differentiation examples

Example 2 Find $y^{\prime}$ for $(y+3)^4=x^7-3 \sin x+y$.
Solution (1) We begin by differentiating both sides with respect to $x$ :
\begin{aligned} \frac{d}{d x}\left((y+3)^4\right) &=\frac{d}{d x}\left(x^7-3 \sin x+y\right) \ 4(y+3)^3 \cdot 1 \cdot y^{\prime} &=7 x^6-3 \cos x+y^{\prime} \end{aligned}
(2) Next we solve for $y^{\prime}$. We start by moving terms involving $y^{\prime}$ to one side of the equation and terms not involving $y^{\prime}$ to the other side:
$$4(y+3)^3 y^{\prime}-y^{\prime}=7 x^6-3 \cos x$$
If necessary (if $y^{\prime}$ is in more than one term), we can factor out $y^{\prime}$ :
$$y^{\prime}\left(4(y+3)^3-1\right)=7 x^6-3 \cos x$$
Finally, we divide to isolate $y^{\prime}$ :
$$y^{\prime}=\frac{7 x^6 \quad 3 \cos x}{4(y+3)^3-1}$$
Because $y^{\prime}$ is all that was requested, there is no third step.
Reading Exercise 25 Find $y^{\prime}$ for $(y+1)^2=x^2+4$
Recall that we are thinking of $y$ as defining a function of $x$ implicitly. Just as both $x \sin x$ and $x \sqrt{x^3-4 x+7}$ are products and require the product rule to differentiate, so does $x \cdot y$.

# 微积分代考

## 数学代写|微积分代写Calculus代写|Differentiating implicitly defined functions

$$\frac{d}{d x} x=1,$$

$$\frac{d}{d x} y=\frac{d y}{d x}=y^{\prime}$$

## 数学代写|微积分代写Calculus代写|Implicit differentiation examples

$$\frac{d}{d x}\left((y+3)^4\right)=\frac{d}{d x}\left(x^7-3 \sin x+y\right) 4(y+3)^3 \cdot 1 \cdot y^{\prime} \quad=7 x^6-3 \cos x+y^{\prime}$$
(2) 接下来我们求解 $y^{\prime}$. 我们首先移动涉及的术语 $y^{\prime}$ 方程的一侧和不涉及的项 $y^{\prime}$ 到另一边:
$$4(y+3)^3 y^{\prime}-y^{\prime}=7 x^6-3 \cos x$$

$$y^{\prime}\left(4(y+3)^3-1\right)=7 x^6-3 \cos x$$

$$y^{\prime}=\frac{7 x^6 \quad 3 \cos x}{4(y+3)^3-1}$$

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