Yukina said:
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
