Implicitly differentiate with respect to \(\displaystyle x \) - This means solve for \(\displaystyle y'\).
Remember to differentiate everything, but only the differentiated y stuff, has y' connected to it.
1.
\(\displaystyle x^{2}y^{2} + x\sin(y) = 1\)
\(\displaystyle [(2x)(y^{2}) + (2yy')(x^{2})] + [(1)(\sin(y)) + (\cos(y)y')(x)] = 0\) Do product rule for terms on the left of the equation. On the right the derivative of a constant is \(\displaystyle 0\).
\(\displaystyle (2yy')(x^{2}) + (\cos(y)y')(x) + (2x)(y^{2}) + (1)(\sin(y)) = 0\) - Group y' containing stuff together.
\(\displaystyle (2yy')(x^{2}) + (\cos(y)y')(x)] = - (2x)(y^{2}) - (1)(\sin(y)) \) - Move terms that don't have y' connected to them on the other side of the equation.
\(\displaystyle y'[(2y)(x^{2}) + (\cos(y))(x)] = -(2x)(y^{2}) - (1)(\sin(y)) = 0\) - Factor out y'.
\(\displaystyle y' = \dfrac{- (2x)(y^{2}) - (1)(\sin(y)) }{ (2y)(x^{2}) + (\cos(y))(x)}\) - Isolate y'
\(\displaystyle y' = \dfrac{- (2x)(y^{2}) - (\sin(y)) }{ (2y)(x^{2}) + (\cos(y))(x)}\) - Sometimes the answer doesn't show the 1 values.
The answer might be written like this, which is the same thing:
\(\displaystyle y' = \dfrac{- (2x)(y^{2}) - (\sin(y)) }{ (2x^{2})(y) + (x)( \cos(y))}\)
2.
\(\displaystyle x^{8}(x + y) = y^{2}(9x - y)\)
\(\displaystyle (8x^{7})(x + y) + (1 + (1)y')(x^{8}) = (2yy')(9x - y) + (9 - (1)y')(y^{2})\) - Use the product rule for terms, some of which are contained in parenthesis.
\(\displaystyle (1 + (1)y')(x^{8}) - (2yy')(9x - y) - (9 - (1)y')(y^{2}) = -(8x^{7})(x + y)\) Group y' terms on the left. The non y' stuff on the right.
\(\displaystyle y'[ (1 + (1))(x^{8}) - (2y)(9x - y) - (9 - (1))(y^{2})] = -(8x^{7})(x + y)\) Factor out y'
\(\displaystyle y' = \dfrac{ -(8x^{7})(x + y) }{ (1 + (1))(x^{8}) - (2y)(9x - y) - (9 - (1))(y^{2}) } \) Isolate y'
\(\displaystyle y' = \dfrac{ -(8x^{7})(x + y) }{ (2x^{8}) - 18xy + 2y) - 8(y^{2}) } \) Might be simplifed like this.
