2 questions (Implicit equation)

[Yukina]

1) regard y as the independent variable an x as the dependent variable and use implicit differentiation to find dx/dy : y sec(x) = x tan(y)
2)If x^2 +xy+y^3=1, find y''' at the point where x=1.

Can anyone help me,please?

 

[galactus]

1) regard y as the independent variable an x as the dependent variable and use implicit differentiation to find dx/dy : y sec(x) = x tan(y)

Use the product rule. When you find the derivative of a x terms, attach a dx/dy.

The derivative of ysec(x) is $\displaystyle ysec(x)tan(x)\frac{dx}{dy}+sec(x)$

The derivative of xtan(y) is $\displaystyle xsec^{2}(y)+tan(y)\frac{dx}{dy}$

$\displaystyle ysec(x)tan(x)\frac{dx}{dy}+sec(x)=xsec^{2}(y)+tan(y)\frac{dx}{dy}$

Now, solve for dx/dy.

2)If x^2 +xy+y^3=1, find y''' at the point where x=1.

This one is rather involved and tedious.

Find y':

$\displaystyle 2x+xy'+y+3y^{2}=0$

$\displaystyle y'=\frac{-2x-y}{x+3y^{2}}$

Now, to find y'', differentiate this. Then, resub y' from before.

Then, differentiate again to find $\displaystyle y'''$. Resub y''.

This is what you're shooting for:

$\displaystyle y'''=\frac{6\left(8x^{4}-x^{3}(120y^{2}-36y+1)-x^{2}y(144y^{2}-33y+1)-xy^{3}(108y^{2}+9y-4)-3y^{5}(9y-1)\right)}{(x+3y^{2})^{5}}$

WHEW!!!. Then, sub in x=1 and y=0