## 数学代写|微积分代写Calculus代写|Second derivatives implicitly

Example 5 Find $y^{\prime \prime}$ for $y^3=4 \sin x-y^2+7$
Solution The usual procedure for finding the second derivative is to start by finding the first derivative. We begin by differentiating each side with respect to $x$ :
\begin{aligned} \frac{d}{d x}\left(y^3\right) &=\frac{d}{d x}\left(4 \sin x-y^2+7\right) \ 3 y^2 \cdot y^{\prime} &=4 \cos x-2 y \cdot y^{\prime} \end{aligned}
Next we solve for $y^{\prime}$ :
\begin{aligned} 3 y^2 y^{\prime}+2 y y^{\prime} &=4 \cos x \ y^{\prime}\left(3 y^2+2 y\right) &=4 \cos x \ y^{\prime} &=\frac{4 \cos x}{3 y^2+2 y} . \end{aligned}
We wish to find the second derivative. Differentiating both sides again with respect to $x$ gives \begin{aligned} \frac{d}{d x} y^{\prime} &=\frac{d}{d x}\left(\frac{4 \cos x}{3 y^2+2 y}\right) \ y^{\prime \prime}-& \frac{\left(3 y^2+2 y\right) \cdot 4(-\sin x)-(4 \cos x)\left(6 y \cdot y^{\prime}+2 \cdot y^{\prime}\right)}{\left(3 y^2+2 y\right)^2} \end{aligned}
There is no need to solve for $y^{\prime \prime}$. However, what we have now is listed in terms of $x, y$, and $y^{\prime}$. To have an expression for $y^{\prime \prime}$ that depends on $x$ and $y$ only, we can replace $y^{\prime}$ with its previous expression:
$$y^{\prime \prime}=\frac{\left(3 y^2+2 y\right) \cdot 4(-\sin x)-(4 \cos x)\left(6 y \cdot \frac{4 \cos x}{3 y^2+2 y}+2 \cdot \frac{4 \cos x}{3 y^2+2 y}\right)}{\left(3 y^2+2 y\right)^2} .$$
The solution can be simplified if desired.
Reading Exercise 27 Find $y^{\prime \prime}$ given that $y^{\prime}=\frac{x^2}{3 x+y^2}$.

## 数学代写|微积分代写Calculus代写|Making implicit explicit

Sometimes an equation in $x$ and $y$ can be solved explicitly for $y$ (more generally, solved for the dependent variable in terms of the independent variable) to avoid implicit differentiation. Let’s reprise example 1 to illustrate.

Example 6 Find the slope of the tangent line to the curve $x^2+y^2=1$ at the point $\left(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right)$.
Solution We begin by solving the equation for $y$ (see figure 5):
\begin{aligned} x^2+y^2 &=1, \ y^2 &=1-x^2, \ y &=\pm \sqrt{1-x^2} . \end{aligned}
Notice that this gives not one, but two, functions (or more; see the earlier discussion). The top half of the curve has positive $y$-coordinates and is therefore represented by $y=\sqrt{1-x^2}$, whereas the bottom half of the curve has negative $y$-coordinates and is represented by $y=-\sqrt{1-x^2}$ (figure 6).

Consulting the graph of the equation (see figure 5), we see that the desired point of tangency lies on the bottom half of the curve, so we use $y=-\sqrt{1-x^2}$ and find the equation of the tangent line to that curve at $x=\frac{1}{\sqrt{2}}$. First we find $y^{\prime}$ :
$$y^{\prime}=-\frac{1}{2}\left(1-x^2\right)^{-\frac{1}{2}} \cdot(-2 x)=\frac{x}{\sqrt{1-x^2}} .$$
Then, we evaluate the derivative at $x=\frac{1}{\sqrt{2}}$ :
$$y^{\prime}\left(\frac{1}{\sqrt{2}}\right)=\frac{\frac{1}{\sqrt{2}}}{\sqrt{1-\frac{1}{2}}}=\frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}=1 .$$
The slope of the tangent line is one, matching the solution to example 1.

The problem with using the method of example 6 is that the first step, solving for $y$, can be difficult or even impossible. Example 3 can be worked in this manner, but it is more difficult. Solving for $y$ in example 5 is very difficult, so that the messy algebra of implicit differentiation is still preferred. Solving for $y$ in example 2 is so difficult that it takes two printed pages just to present the solution, and then we would still need to take the derivative of that mess. And solving for $y$ in example 4 cannot be done at all. Implicit differentiation is, by far, the preferred method.

# 微积分代考

## 数学代写|微积分代写Calculus代写|Second derivatives implicitly

$$\frac{d}{d x}\left(y^3\right)=\frac{d}{d x}\left(4 \sin x-y^2+7\right) 3 y^2 \cdot y^{\prime}=4 \cos x-2 y \cdot y^{\prime}$$

$$3 y^2 y^{\prime}+2 y y^{\prime}=4 \cos x y^{\prime}\left(3 y^2+2 y\right) \quad=4 \cos x y^{\prime}=\frac{4 \cos x}{3 y^2+2 y} .$$

$$\frac{d}{d x} y^{\prime}=\frac{d}{d x}\left(\frac{4 \cos x}{3 y^2+2 y}\right) y^{\prime \prime}-\quad \frac{\left(3 y^2+2 y\right) \cdot 4(-\sin x)-(4 \cos x)\left(6 y \cdot y^{\prime}+2 \cdot y^{\prime}\right)}{\left(3 y^2+2 y\right)^2}$$

$$y^{\prime \prime}=\frac{\left(3 y^2+2 y\right) \cdot 4(-\sin x)-(4 \cos x)\left(6 y \cdot \frac{4 \cos x}{3 y^2+2 y}+2 \cdot \frac{4 \cos x}{3 y^2+2 y}\right)}{\left(3 y^2+2 y\right)^2} .$$

## 数学代写|微积分代写Calculus代写|Making implicit explicit

$$x^2+y^2=1, y^2 \quad=1-x^2, y=\pm \sqrt{1-x^2} .$$

$$y^{\prime}=-\frac{1}{2}\left(1-x^2\right)^{-\frac{1}{2}} \cdot(-2 x)=\frac{x}{\sqrt{1-x^2}} .$$

$$y^{\prime}\left(\frac{1}{\sqrt{2}}\right)=\frac{\frac{1}{\sqrt{2}}}{\sqrt{1-\frac{1}{2}}}=\frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}=1 .$$

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