derivative of y=sec(xy^3) ?!!
y= sec x + sec y^3 ??
anyone help me out please
The part that I highlighted in red is correct. Remember that we used implicit differentiation. Suppose I want to differentiate the equation:
$\displaystyle y=3x^2+y^3$
I have to use implicit differentiation:
$\displaystyle y'=6x+3y^2y'$
Notice that $\displaystyle y'$ is on both sides of the equation. You can often solve for $\displaystyle y'$.
$\displaystyle y'+-3y^2y'=6x$
$\displaystyle y'(1-3y^2)=6x$
$\displaystyle y'=\frac{6x}{1-3y^2}$
With your equation, we can do the same thing:
$\displaystyle y'=sec(xy^3)][tan(xy^3)][y^3 + 3xy^2y']$
$\displaystyle y'-3xy^2y'sec(xy^3)tan(xy^3)=y^3sec(xy^3)tan(xy^3)$
$\displaystyle y'[1-3xy^2sec(xy^3)tan(xy^3)]=y^3sec(xy^3)tan(xy^3)$
$\displaystyle y'=\frac{y^3sec(xy^3)tan(xy^3)}{1-3xy^2sec(xy^3)tan(xy^3)}$
You have to EXPAND the right side of this equation.
$\displaystyle y'=sec(xy^3)][tan(xy^3)][y^3 + 3xy^2y']$
$\displaystyle y'=y^3sec(xy^3)tan(xy^3)+3xy^2y'sec(xy^3)tan(xy^3)$
$\displaystyle y'-3xy^2y'sec(xy^3)tan(xy^3)=y^3sec(xy^3)tan(xy^3)$
$\displaystyle y'[1-3xy^2sec(xy^3)tan(xy^3)]=y^3sec(xy^3)tan(xy^3)$
$\displaystyle y'=\frac{y^3sec(xy^3)tan(xy^3)}{1-3xy^2sec(xy^3)tan(xy^3)}$
When you expand the right side of the equation: $\displaystyle y'=sec(xy^3)][tan(xy^3)][y^3 + 3xy^2y']$ you use the distributive property of algebra. To make this more obvious, let $\displaystyle A=sec(xy^3)tan(xy^3)$. So the equation becomes:
$\displaystyle y'=A(y^3+3xy^2y')=Ay^3+A3xy^2y'$
Then I subtracted the $\displaystyle A3xy^2y'$ from both sides and factored the left side to isolate $\displaystyle y'$.