Question

11. Find \(\frac{{dy}}{{dx}}\) by implicit differentiation.

11. \(\sin x + \cos y = 2x - 3y\)

Short answer

The value is \(\frac{{dy}}{{dx}} = \frac{{2 - \cos x}}{{3 - \sin y}}\).

Step-by-step solution

Differentiate the given equation

The given equation is\(\sin x + \cos y = 2x - 3y\). Both sides of this equation are differentiated with respect to\(x\)as follows:

\(\begin{aligned}\frac{d}{{dx}}\left( {\sin x + \cos y} \right) &= \frac{d}{{dx}}\left( {2x - 3y} \right)\\\cos x - \sin y \cdot y' &= 2 - 3y'\end{aligned}\)

Simplify the resulting equation

Simplify the resulting equation to obtain the expression for\(\frac{{dy}}{{dx}}\), as shown below:

\(\begin{aligned}3y' - \sin y \cdot y' &= 2 - \cos x\\y'\left( {3 - \sin y} \right) &= 2 - \cos x\\y' &= \frac{{2 - \cos x}}{{3 - \sin y}}\end{aligned}\).

Thus, the value is \(\frac{{dy}}{{dx}} = \frac{{2 - \cos x}}{{3 - \sin y}}\).