The value of dy/dx for the functions are
(i) [tex]\frac{dy}{dx} = 4x^{2}sin2x.cos2x+ 2x. sin^{2}2x[/tex]
(ii) [tex]\frac{dy}{dx} =\frac{- y(1+3x^{2})}{2x(1+x^{2}) }[/tex]
Differentiation
From the question, we are to determine dy/dx for the given functions
(i) [tex]x^{2} sin^{2}2x[/tex]
Let [tex]u = x^{2}[/tex]
and
[tex]v = sin^{2} 2x[/tex]
Also,
Let [tex]w= sin2x[/tex]
∴ [tex]v = w^{2}[/tex]
First, we will determine [tex]\frac{dv}{dx}[/tex]
Using the Chain rule
[tex]\frac{dv}{dx} = \frac{dv}{dw}.\frac{dw}{dx}[/tex]
[tex]v = w^{2}[/tex]
∴ [tex]\frac{dv}{dw} =2w[/tex]
Also,
[tex]w= sin2x[/tex]
∴ [tex]\frac{dw}{dx} =2cos2x[/tex]
Thus,
[tex]\frac{dv}{dx} = 2w \times 2cos2x[/tex]
[tex]\frac{dv}{dx} = 2sin2x \times 2cos2x[/tex]
[tex]\frac{dv}{dx} = 4sin2x . cos2x[/tex]
Now, using the product rule
[tex]\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}[/tex]
From above
[tex]u = x^{2}[/tex]
∴ [tex]\frac{du}{dx}=2x[/tex]
Now,
[tex]\frac{dy}{dx} = x^{2} (4sin2x.cos2x)+ sin^{2}2x (2x)[/tex]
[tex]\frac{dy}{dx} = 4x^{2}sin2x.cos2x+ 2x. sin^{2}2x[/tex]
(ii) [tex]xy^{2}+y^{2}x^{3} +2=0[/tex]
Then,
[tex]x.2y\frac{dy}{dx}+ y^{2}(1)+y^{2}.3x^{2} + x^{3}.2y\frac{dy}{dx} +0=0[/tex]
[tex]2xy\frac{dy}{dx}+ y^{2}+3x^{2}y^{2} + 2x^{3}y\frac{dy}{dx} =0[/tex]
[tex]2xy\frac{dy}{dx}+2x^{3}y\frac{dy}{dx} =- y^{2}-3x^{2}y^{2}[/tex]
[tex]\frac{dy}{dx} (2xy+2x^{3}y)=- y^{2}(1+3x^{2})[/tex]
[tex]\frac{dy}{dx} =\frac{- y^{2}(1+3x^{2})}{2xy+2x^{3}y}[/tex]
[tex]\frac{dy}{dx} =\frac{- y^{2}(1+3x^{2})}{2xy(1+x^{2}) }[/tex]
[tex]\frac{dy}{dx} =\frac{- y(1+3x^{2})}{2x(1+x^{2}) }[/tex]
Hence, the value of dy/dx for the functions are
(i) [tex]\frac{dy}{dx} = 4x^{2}sin2x.cos2x+ 2x. sin^{2}2x[/tex]
(ii) [tex]\frac{dy}{dx} =\frac{- y(1+3x^{2})}{2x(1+x^{2}) }[/tex]
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