Question

Find $\frac{dy}{dx}$ if
(a) $x^3+2x^{2}y+3x{y}^2+4y^3=5$
(b) $x=2{\cos }^3\theta ,y=2{\sin }^3\theta$
(c) $y={\sin }^{-1}\left(2x\sqrt{1-x^2}\right);-1\le x\le 1$

Answer 1

(a) $x^3+2x^{2}y+3x{y}^2+4y^3=5$

Differentiating both the sides wrt x

$\Rightarrow$ ${3x}^2+2x^2\frac{dy}{dx}+4xy+6xy\frac{dy}{dx}+3y^2+12y^2\frac{dy}{dx}=0$

$\Rightarrow$ $\frac{dy}{dx}(2x^2+6xy+12y^2){=-(3x}^2+4xy+3y^2)$

$\Rightarrow$ $\frac{dy}{dx}=\frac{{-(3x}^2+4xy+3y^2)}{(2x^2+6xy+12y^2)}$

(b) $x=2{cos}^3\theta ,y=2{sin}^3\theta$

$\Rightarrow$ $\frac{dy}{d\theta }=6{sin}^2\theta cos\theta ,\frac{dx}{d\theta }=-6{cos}^2\theta sin\theta$

$\Rightarrow$ $\frac{dy}{dx}=-tan\theta$

(c) $y={sin}^{-1}\left(2x\sqrt{1-x^2}\right);-1\le x\le 1$

Let $x=sin\theta$, $\frac{dx}{d\theta }=cos\theta$

$\Rightarrow$ $y={sin}^{-1}\left(2sin\theta \sqrt{1-{sin}^2\theta }\right)$

$\Rightarrow$ $y={sin}^{-1}\left(2sin\theta cos\theta \right)$

$\Rightarrow$ $y={sin}^{-1}\left(sin2\theta \right)\because sin2\theta =2sin\theta cos\theta$

$\Rightarrow$ $y=2\theta$

$\Rightarrow$ $\frac{dy}{dx}=2\frac{d\theta }{dx}=2sec\theta =2sec\left({sin}^{-1}x\right)$