Question

Question: 12. Find \(\frac{{dy}}{{dx}}\) by implicit differentiation.

12. \({e^x}\sin y = x + y\)

Short answer

The value is \(\frac{{dy}}{{dx}} = \frac{{1 - {e^x}\sin y}}{{{e^x}\cos y - 1}}\).

Step-by-step solution

Differentiate the given equation

The given equation is\({e^x}\sin y = x + y\). Both sides of this equation are differentiated with respect to\(x\)as follows:

\(\begin{aligned}\frac{d}{{dx}}\left( {{e^x}\sin y} \right) &= \frac{d}{{dx}}\left( {x + y} \right)\\{e^x}\cos y \cdot y' + \sin y \cdot {e^x} &= 1 + y'\end{aligned}\)

Simplify the resulting equation 

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

\(\begin{aligned}{e^x}\cos y \cdot y' - y' &= 1 - {e^x}\sin y\\y'\left( {{e^x}\cos y - 1} \right) &= 1 - {e^x}\sin y\\y' &= \frac{{1 - {e^x}\sin y}}{{{e^x}\cos y - 1}}\end{aligned}\).

Thus, the value is \(\frac{{dy}}{{dx}} = \frac{{1 - {e^x}\sin y}}{{{e^x}\cos y - 1}}\).