Question

$${ y }^{ 2 } +y \cos x=y \sin xdifferntiate$$

Answer

$$d=-(y+cos(x))/(e^2*f^2*r*n*t^2*a*sin(x))$$

Solution


Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[{y}^{2}+y\cos{x}=y(\sin{x})d{\imath }^{2}{f}^{2}{e}^{2}rn{t}^{2}a\]
Use Square Rule: \({i}^{2}=-1\).
\[{y}^{2}+y\cos{x}=y(\sin{x})d\times -1\times {f}^{2}{e}^{2}rn{t}^{2}a\]
Simplify  \(y(\sin{x})d\times -1\times {f}^{2}{e}^{2}rn{t}^{2}a\)  to  \(y(\sin{x})d\times -{f}^{2}{e}^{2}rn{t}^{2}a\).
\[{y}^{2}+y\cos{x}=y(\sin{x})d\times -{f}^{2}{e}^{2}rn{t}^{2}a\]
Regroup terms.
\[{y}^{2}+y\cos{x}=-{e}^{2}yd{f}^{2}rn{t}^{2}a\sin{x}\]
Factor out the common term \(y\).
\[y(y+\cos{x})=-{e}^{2}yd{f}^{2}rn{t}^{2}a\sin{x}\]
Cancel \(y\) on both sides.
\[y+\cos{x}=-{e}^{2}d{f}^{2}rn{t}^{2}a\sin{x}\]
Divide both sides by \(-{e}^{2}\).
\[-\frac{y+\cos{x}}{{e}^{2}}=d{f}^{2}rn{t}^{2}a\sin{x}\]
Divide both sides by \({f}^{2}\).
\[-\frac{\frac{y+\cos{x}}{{e}^{2}}}{{f}^{2}}=drn{t}^{2}a\sin{x}\]
Simplify  \(\frac{\frac{y+\cos{x}}{{e}^{2}}}{{f}^{2}}\)  to  \(\frac{y+\cos{x}}{{e}^{2}{f}^{2}}\).
\[-\frac{y+\cos{x}}{{e}^{2}{f}^{2}}=drn{t}^{2}a\sin{x}\]
Divide both sides by \(r\).
\[-\frac{\frac{y+\cos{x}}{{e}^{2}{f}^{2}}}{r}=dn{t}^{2}a\sin{x}\]
Simplify  \(\frac{\frac{y+\cos{x}}{{e}^{2}{f}^{2}}}{r}\)  to  \(\frac{y+\cos{x}}{{e}^{2}{f}^{2}r}\).
\[-\frac{y+\cos{x}}{{e}^{2}{f}^{2}r}=dn{t}^{2}a\sin{x}\]
Divide both sides by \(n\).
\[-\frac{\frac{y+\cos{x}}{{e}^{2}{f}^{2}r}}{n}=d{t}^{2}a\sin{x}\]
Simplify  \(\frac{\frac{y+\cos{x}}{{e}^{2}{f}^{2}r}}{n}\)  to  \(\frac{y+\cos{x}}{{e}^{2}{f}^{2}rn}\).
\[-\frac{y+\cos{x}}{{e}^{2}{f}^{2}rn}=d{t}^{2}a\sin{x}\]
Divide both sides by \({t}^{2}\).
\[-\frac{\frac{y+\cos{x}}{{e}^{2}{f}^{2}rn}}{{t}^{2}}=da\sin{x}\]
Simplify  \(\frac{\frac{y+\cos{x}}{{e}^{2}{f}^{2}rn}}{{t}^{2}}\)  to  \(\frac{y+\cos{x}}{{e}^{2}{f}^{2}rn{t}^{2}}\).
\[-\frac{y+\cos{x}}{{e}^{2}{f}^{2}rn{t}^{2}}=da\sin{x}\]
Divide both sides by \(a\).
\[-\frac{\frac{y+\cos{x}}{{e}^{2}{f}^{2}rn{t}^{2}}}{a}=d\sin{x}\]
Simplify  \(\frac{\frac{y+\cos{x}}{{e}^{2}{f}^{2}rn{t}^{2}}}{a}\)  to  \(\frac{y+\cos{x}}{{e}^{2}{f}^{2}rn{t}^{2}a}\).
\[-\frac{y+\cos{x}}{{e}^{2}{f}^{2}rn{t}^{2}a}=d\sin{x}\]
Divide both sides by \(\sin{x}\).
\[-\frac{\frac{y+\cos{x}}{{e}^{2}{f}^{2}rn{t}^{2}a}}{\sin{x}}=d\]
Simplify  \(\frac{\frac{y+\cos{x}}{{e}^{2}{f}^{2}rn{t}^{2}a}}{\sin{x}}\)  to  \(\frac{y+\cos{x}}{{e}^{2}{f}^{2}rn{t}^{2}a\sin{x}}\).
\[-\frac{y+\cos{x}}{{e}^{2}{f}^{2}rn{t}^{2}a\sin{x}}=d\]
Switch sides.
\[d=-\frac{y+\cos{x}}{{e}^{2}{f}^{2}rn{t}^{2}a\sin{x}}\]