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A curve has equation $$x^3-xy+y^2=7$$
a) show that tangent to the curve is horizontal at $(1,3)$
b) determine $y''$ at $(1,3)$

Solution for a)
using product rule, I got this,

$$3x^2(y+y^2)=0$$ However the teacher got,
$$**3x^2-(y+xy')+2yy'=0**$$ then sub $x=1,y=3$ $$3-(3+y')+6y'=0$$ $$y'(1)=0$$

Solution for b)

$$(3x^2-xy'-y+2yy')=0$$ $$**6x-y'-xy''-y'+2(y')^2+2yy''=0**$$ $$6x-2y'+y''(2y-x)+2(y')^2=0$$ sub x=1,y=3 $$6+y''(1)(6-1)=0$$ using y'(1)=0 $$y''(1)=-\frac{6}{5}$$

I have difficult understanding steps where the stars are marked!

Appreciate your helps! Thanks

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2 Answers 2

up vote 3 down vote accepted

Differentiate both sides with respect to $x$. Recall that $y$ is a function of $x$. Maybe it will help if instead of $y$ we write $f(x)$. So your equation becomes $$x^3-xf(x)+(f(x))^2=7.$$ Now let's differentiate. The derivative of $x^3$ is $3x^2$.

Now differentiate $xf(x)$. This is a product. So we use the Product Rule, and get $xf'(x)+f(x)$.

Now differentiate the function $(f(x))^2$. You can view this as a product. Then the derivative is $f(x)f'(x)+f(x)f'(x)=2f(x)f'(x)$. But for more complicated expressions, it is better to use the Chain Rule. Our function $(f(x))^2$ is $h(f(x))$ where $h(u)=u^2$. So the derivative is $(2u)f'(x)$, that is, $2f(x)f'(x)$.

On the right-hand side, we have the constant function $7$. Its derivative is $0$. So we get $$3x^2 -(xf'(x)+f(x))+2f(x)f'(x)=0.\tag{$1$}$$ Now we go back to the letter $y$. This really should be $y(x)$, but we suppress the $x$. Look back at Equation $(1)$, and write $y$ for $f(x)$ and $y'$ for $f'(x)$. We get $$3x^2-(xy'+y)+2yy'=0.$$

Now let's do it without the replacement by $f(x)$. The derivative of $x^3$ is $3x^2$. For the derivative of $xy$ with respect to $x$, note that $xy$ is the product of two functions. So we use the Product rule and get $xy'+y$. Finally, to differentiate $y^2$ with respect to $x$, note that by the Chain Rule, $\frac{d}{dx}(y^2)=\frac{d}{dy}(y^2)\frac{dy}{dx}=2y\frac{dy}{dx}=2yy'$.

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Could you please collaborate more $−xy$ partial: $−y−xy′$ –  Sb Sangpi Mar 7 '13 at 11:50
1  
There is no partial derivative involved, $y$ is a function of $x$ and we are differentiating with respect to $x$. The function $xy$ is the product of two functions, $x$ and $y$. To differentiate, use product rule: first, times derivative of second, plus second, times derivative of first. That's $(x)(y')+(y)(1)=xy'+y$. Similarly, the derivative of $x^6y$ would be $x^6y'+(y)(6x^5)$. –  André Nicolas Mar 7 '13 at 12:03

Hint:

Take the derivative of each term by itself using partial derivatives:

$$x^3-xy+y^2=7$$

$x^3 ~\text{partial}~: 3x^2$ (Power Rule)

$-xy ~\text{partial}~: -y -xy'$ (Product Rule)

$y^2 ~\text{partial}~: 2 y y'$ (Power Rule)

Now sum those and you get what the teacher has.

Find partial of:

$$(3x^2-xy'-y+2yy')=0$$

Again, do the partial of one term at a time:

$3x^2 ~\text{partial}~: 6x$

$-xy' ~\text{partial}~: -y' -xy''$

$-y ~\text{partial}~: -y'$

$2yy' ~\text{partial}~: 2(y')(y') + 2yy''$

Do you see your issues now?

Regards

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Could you please collaborate more $-xy$ partial: $-y-xy'$ –  Sb Sangpi Mar 7 '13 at 10:38
    
@SbSangpi: It looks like Andre' provided a nice response to your follow-up. Regards –  Amzoti Mar 7 '13 at 15:14
    
Nice work, Amzoti! –  amWhy Apr 24 '13 at 0:42

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