If cosx2y3+sinx2y3=k then dydx

A
$- \frac{2 y}{3 x}$
B
$- \frac{3 x}{2 y}$
C
$\frac{(- \sin (x^{2} y^{3}) + ({cosx}^{2} y^{3})) 2 y}{3 x}$
D
$0$

Solution:

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

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

$[- \sin (x^{2} y^{3}) + \cos (x^{2} y^{3})] [2 x y^{3} + 3 x^{2} y^{2} \frac{d y}{d x}] = 0$

It implies that

$\begin{array}{rcl} 2 x y^{3} + 3 x^{2} y^{2} \frac{d y}{d x} & = & 0 \\ 2 y + 3 x \frac{d y}{d x} & = & 0 \\ \frac{d y}{d x} & = & - \frac{2 y}{3 x} \end{array}$

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

Solution:

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