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Question Number 121729 by oustmuchiya@gmail.com last updated on 11/Nov/20

	Answered by bemath last updated on 11/Nov/20

	$(i) 2 {y y}^{'} + 3 x^{2} - 3 y^{2} y^{'} = 3 y^{'}$ $\Leftrightarrow 3 x^{2} = y^{'} (3 + 3 y^{2} - 2 y)$ $\Leftrightarrow y^{'} = \frac{3 x^{2}}{3 y^{2} - 2 y + 3}$
	Answered by bemath last updated on 11/Nov/20

	$(i i) y^{'} . \cos y + 2 {x y}^{3} + 3 x^{2} y^{2} . y^{'} + \sin x = 3 y^{'}$ $\Rightarrow 2 {x y}^{3} + \sin x = y^{'} (3 - 3 x^{2} y^{2} - \cos y)$ $\Leftrightarrow y^{'} = \frac{2 {x y}^{3} + \sin x}{3 - 3 x^{2} y^{2} - \cos y} .$
	Answered by bemath last updated on 11/Nov/20

	$(i i i) 2 y^{3} - x \cos y + \frac{1}{2} x^{- 1} y^{2} = 7$ $\Rightarrow 6 y^{2} . y^{'} - (\cos y - {x y}^{'} \sin y) + \frac{1}{2} (- x^{- 2} y^{2} + 2 x^{- 1} y . y^{'}) = 0$ $\Rightarrow 6 y^{2} . y^{'} - \cos y + {x y}^{'} \sin y - \frac{y^{2}}{2 x^{2}} + \frac{y . y^{'}}{x} = 0$ $\Rightarrow (6 y^{2} + x \sin y + \frac{y}{x}) . y^{'} = \frac{y^{2}}{2 x^{2}} + \cos y$ $\Rightarrow (12 x^{2} y^{2} + 2 x^{3} \sin y + 2 x y) . y^{'} = y^{2} + 2 x^{2} \cos y$ $y^{'} = \frac{y^{2} + 2 x^{2} \cos y}{12 x^{2} y^{2} + 2 x^{3} \sin y + 2 x y} .$
	Answered by bemath last updated on 11/Nov/20
	$(iv) using product rule let { ((u=(5x−3)^4 ⇒u′=20(5x−3)^3 )),((v=3(3x)^(1/3) ⇒v′= 3(3x)^(−(2/3)) )) :} y′=u′v+uv′$
	$(i v) u s i n g p r o d u c t r u l e$ $l e t {\begin{cases} u = {(5 x - 3)}^{4} \Rightarrow u^{'} = 20 {(5 x - 3)}^{3} \\ v = 3 {(3 x)}^{\frac{1}{3}} \Rightarrow v^{'} = 3 {(3 x)}^{- \frac{2}{3}} \end{cases}$ $y^{'} = u^{'} v + {u v}^{'}$
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