Question

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

15.\(y\cos x = {x^2} + {y^2}\)

Short answer

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

Step-by-step solution

Differentiate the given equation

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

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

Simplify the resulting equation

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

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

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