Question

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

\(2x{e^y} + y{e^x} = 3\)

Short answer

The value is\(\frac{{dy}}{{dx}} = - \frac{{2{e^y} + y{e^x}}}{{2x{e^y} + {e^x}}}\).

Step-by-step solution

Differentiate the given equation

The given equation is\(2x{e^y} + y{e^x} = 3\). Both sides of this equation are differentiated with respect to\(x\)as follows:

\(\begin{aligned}\frac{d}{{dx}}\left( {2x{e^y} + y{e^x}} \right) &= \frac{d}{{dx}}\left( 3 \right)\\2x \cdot {e^y}y' + {e^y} \cdot 2 + \left( {y{e^x} + {e^x}y'} \right) &= 0\end{aligned}\)

Simplify the resulting equation

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

\(\begin{aligned}2x{e^y}y' + {e^x}y' &= - 2{e^y} - y{e^x}\\y'\left( {2x{e^y} + {e^x}} \right) &= - \left( {2{e^y} + y{e^x}} \right)\\y' &= - \frac{{2{e^y} + y{e^x}}}{{2x{e^y} + {e^x}}}\end{aligned}\).

Thus, the value is \(\frac{{dy}}{{dx}} = - \frac{{2{e^y} + y{e^x}}}{{2x{e^y} + {e^x}}}\).