Question

Please help me answer this question

Answer 1

The value of dy/dx for the functions are

(i) $\frac{dy}{dx} = 4x^{2}sin2x.cos2x+ 2x. sin^{2}2x$

(ii) $\frac{dy}{dx} =\frac{- y(1+3x^{2})}{2x(1+x^{2}) }$

Differentiation

From the question, we are to determine dy/dx for the given functions

(i) $x^{2} sin^{2}2x$

Let $u = x^{2}$

and

$v = sin^{2} 2x$

Also,

Let $w= sin2x$

∴ $v = w^{2}$

First, we will determine $\frac{dv}{dx}$

Using the Chain rule
$\frac{dv}{dx} = \frac{dv}{dw}.\frac{dw}{dx}$

$v = w^{2}$

∴ $\frac{dv}{dw} =2w$

Also,

$w= sin2x$

∴ $\frac{dw}{dx} =2cos2x$

Thus,

$\frac{dv}{dx} = 2w \times 2cos2x$

$\frac{dv}{dx} = 2sin2x \times 2cos2x$

$\frac{dv}{dx} = 4sin2x . cos2x$

Now, using the product rule

$\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$

From above

$u = x^{2}$

∴ $\frac{du}{dx}=2x$

Now,

$\frac{dy}{dx} = x^{2} (4sin2x.cos2x)+ sin^{2}2x (2x)$

$\frac{dy}{dx} = 4x^{2}sin2x.cos2x+ 2x. sin^{2}2x$

(ii) $xy^{2}+y^{2}x^{3} +2=0$

Then,

$x.2y\frac{dy}{dx}+ y^{2}(1)+y^{2}.3x^{2} + x^{3}.2y\frac{dy}{dx} +0=0$

$2xy\frac{dy}{dx}+ y^{2}+3x^{2}y^{2} + 2x^{3}y\frac{dy}{dx} =0$

$2xy\frac{dy}{dx}+2x^{3}y\frac{dy}{dx} =- y^{2}-3x^{2}y^{2}$

$\frac{dy}{dx} (2xy+2x^{3}y)=- y^{2}(1+3x^{2})$

$\frac{dy}{dx} =\frac{- y^{2}(1+3x^{2})}{2xy+2x^{3}y}$

$\frac{dy}{dx} =\frac{- y^{2}(1+3x^{2})}{2xy(1+x^{2}) }$

$\frac{dy}{dx} =\frac{- y(1+3x^{2})}{2x(1+x^{2}) }$

Hence, the value of dy/dx for the functions are

(i) $\frac{dy}{dx} = 4x^{2}sin2x.cos2x+ 2x. sin^{2}2x$

(ii) $\frac{dy}{dx} =\frac{- y(1+3x^{2})}{2x(1+x^{2}) }$

Learn more on Differentiation here: brainly.com/question/24024883

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