If cosx2y3+sinx2y3=k then dydx

# If $\mathrm{cos}\left({\mathrm{x}}^{2}{\mathrm{y}}^{3}\right)+\mathrm{sin}\left({\mathrm{x}}^{2}{\mathrm{y}}^{3}\right)=\mathrm{k}$ then $\frac{\mathrm{dy}}{\mathrm{dx}}$

1. A

$-\frac{2\mathrm{y}}{3\mathrm{x}}$

2. B

$-\frac{3\mathrm{x}}{2\mathrm{y}}$

3. C

$\frac{\left(-\mathrm{sin}\left({\mathrm{x}}^{2}{\mathrm{y}}^{3}\right)+\left({\mathrm{cosx}}^{2}{\mathrm{y}}^{3}\right)\right)2\mathrm{y}}{3\mathrm{x}}$

4. D

$0$

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### Solution:

The given equation is $\mathrm{cos}\left({\mathrm{x}}^{2}{\mathrm{y}}^{3}\right)+\mathrm{sin}\left({\mathrm{x}}^{2}{\mathrm{y}}^{3}\right)=\mathrm{k}$

Use the implicit differentiation, differentiate both sides with respect to $x$

$\left[-\mathrm{sin}\left({x}^{2}{y}^{3}\right)+\mathrm{cos}\left({x}^{2}{y}^{3}\right)\right]\left[2x{y}^{3}+3{x}^{2}{y}^{2}\frac{dy}{dx}\right]=0$

It implies that

$\begin{array}{rcl}2x{y}^{3}+3{x}^{2}{y}^{2}\frac{dy}{dx}& =& 0\\ 2y+3x\frac{dy}{dx}& =& 0\\ \frac{dy}{dx}& =& \overline{)\mathbf{-}\frac{\mathbf{2}\mathbf{y}}{\mathbf{3}\mathbf{x}}}\end{array}$

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