4. Find $$ y^{\prime} $$ for each of the following functions (a) $$ y=e^{x} \sin x $$ (b) $$ y=\ln x \cos x+e^{5 x} $$ (c) $$ y=\frac{x^{2}-1}{\sin x} $$ (d) $$ y=\frac{\ln x}{\cos x} $$ (e) $$ y=\frac{e^{3} x}{x^{2}} $$ 5. Find $$ y^{\prime} $$ for each of the following functions (a) $$ y=\sin 2 x+y^{3}+8 x $$ (b) $$ x^{2}+y^{2}-\ln y+e^{-2 x}-3 x=0 $$ (c) $$ x y+e^{x y}=5 $$

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$$
\textbf{4(a)}\quad y = e^{x}\sin x.
$$

Differentiate using the product rule:

$$
y' = \frac{d}{dx}(e^{x})\sin x + e^{x}\frac{d}{dx}(\sin x) = e^{x}\sin x + e^{x}\cos x = e^{x}(\sin x+\cos x).
$$

$$
\textbf{4(b)}\quad y = \ln x \cos x + e^{5x}.
$$

Differentiate term‐by‐term:

- For $$ \ln x \cos x $$ use the product rule:
  $$
  \frac{d}{dx}(\ln x \cos x) = \frac{1}{x}\cos x + \ln x(-\sin x)=\frac{\cos x}{x}-\ln x\sin x.
  $$
- For $$ e^{5x} $$:
  $$
  \frac{d}{dx}(e^{5x})=5e^{5x}.
  $$

Thus,

$$
y'=\frac{\cos x}{x}-\ln x\sin x+5e^{5x}.
$$

$$
\textbf{4(c)}\quad y = \frac{x^{2}-1}{\sin x}.
$$

Using the quotient rule where $$ u=x^{2}-1 $$ and $$ v=\sin x $$:
$$
y'=\frac{u'v - uv'}{v^{2}},
$$
with $$ u'=2x $$ and $$ v'=\cos x $$. Therefore,
$$
y'=\frac{2x\sin x - (x^{2}-1)\cos x}{\sin^{2} x}.
$$

$$
\textbf{4(d)}\quad y = \frac{\ln x}{\cos x}.
$$

Again, applying the quotient rule with $$ u = \ln x $$ (so $$ u'=\frac{1}{x} $$) and $$ v = \cos x $$ (so $$ v'=-\sin x $$):
$$
y'=\frac{u'v - uv'}{v^{2}} = \frac{\frac{1}{x}\cos x - (\ln x)(-\sin x)}{\cos^{2}x} = \frac{\frac{\cos x}{x}+\ln x\sin x}{\cos^{2}x}.
$$

$$
\textbf{4(e)}\quad y = \frac{e^{3}x}{x^{2}}.
$$

Since $$ e^{3} $$ is a constant, we can simplify:
$$
y = \frac{e^{3}x}{x^{2}} = \frac{e^{3}}{x}.
$$
Differentiate using the power rule:
$$
y = e^{3}x^{-1} \quad \Rightarrow \quad y' = -e^{3}x^{-2} = -\frac{e^{3}}{x^{2}}.
$$

$$
\textbf{5(a)}\quad y = \sin 2x + y^{3} + 8x.
$$

Differentiate both sides with respect to $$ x $$. On the left, $$ \frac{d}{dx}(y) = y' $$. On the right:
$$
\frac{d}{dx}(\sin 2x)=2\cos2x,\quad \frac{d}{dx}(y^{3})=3y^{2}y',\quad \frac{d}{dx}(8x)=8.
$$
Thus,
$$
y' = 2\cos2x+3y^{2}y'+8.
$$
Solve for $$ y' $$ by isolating terms:
$$
y'-3y^{2}y'=2\cos2x+8,
$$
$$
y'(1-3y^{2})=2\cos2x+8,
$$
$$
y'=\frac{2\cos2x+8}{1-3y^{2}}.
$$

$$
\textbf{5(b)}\quad x^{2}+y^{2}-\ln y+e^{-2x}-3x=0.
$$

Differentiate each term with respect to $$ x $$:
$$
\frac{d}{dx}(x^{2})=2x,\quad \frac{d}{dx}(y^{2})=2yy',\quad \frac{d}{dx}(-\ln y)=-\frac{1}{y}y',\quad \frac{d}{dx}(e^{-2x})=-2e^{-2x},\quad \frac{d}{dx}(-3x)=-3.
$$
Thus, the derivative equation is:
$$
2x + 2yy' -\frac{1}{y}y'-2e^{-2x}-3=0.
$$
Group the $$ y' $$-terms:
$$
2yy'-\frac{y'}{y} = y'\left(2y-\frac{1}{y}\right).
$$
So,
$$
2x -2e^{-2x}-3+y'\left(2y-\frac{1}{y}\right)=0.
$$
Isolate $$ y' $$:
$$
y'\left(2y-\frac{1}{y}\right)= -2x+2e^{-2x}+3,
$$
$$
y'=\frac{-2x+2e^{-2x}+3}{2y-\frac{1}{y}}.
$$
Multiplying numerator and denominator by $$ y $$ (if desired) gives:
$$
y'=\frac{-2xy+2ye^{-2x}+3y}{2y^{2}-1}.
$$

$$
\textbf{5(c)}\quad xy+e^{xy}=5.
$$

Differentiate both sides with respect to $$ x $$.

For $$ xy $$:
$$
\frac{d}{dx}(xy)= x y' + y.
$$
For $$ e^{xy} $$, apply the chain rule:
$$
\frac{d}{dx}(e^{xy})= e^{xy}\cdot \frac{d}{dx}(xy)=e^{xy}(x y' + y).
$$
Thus, the differentiated equation is:
$$
x y' + y + e^{xy}(x y' + y)=0.
$$
Factor $$ (xy' + y) $$:
$$
(xy'+ y)(1+e^{xy})=0.
$$
Since $$ 1+e^{xy} \ne 0 $$ for real $$ x,y $$ (because $$ e^{xy}>0 $$), we must have:
$$
xy' + y = 0.
$$
Solving for $$ y' $$:
$$
xy'=-y \quad \Longrightarrow \quad y'=-\frac{y}{x}.
$$
$$ \textbf{4(a)}\quad y' = e^{x}(\sin x + \cos x) $$ $$ \textbf{4(b)}\quad y' = \frac{\cos x}{x} - \ln x \sin x + 5e^{5x} $$ $$ \textbf{4(c)}\quad y' = \frac{2x\sin x - (x^{2} - 1)\cos x}{\sin^{2} x} $$ $$ \textbf{4(d)}\quad y' = \frac{\frac{\cos x}{x} + \ln x \sin x}{\cos^{2} x} $$ $$ \textbf{4(e)}\quad y' = -\frac{e^{3}}{x^{2}} $$ $$ \textbf{5(a)}\quad y' = \frac{2\cos2x + 8}{1 - 3y^{2}} $$ $$ \textbf{5(b)}\quad y' = \frac{-2x + 2e^{-2x} + 3}{2y - \frac{1}{y}} \quad \text{or} \quad y' = \frac{-2xy + 2ye^{-2x} + 3y}{2y^{2} - 1} $$ $$ \textbf{5(c)}\quad y' = -\frac{y}{x} $$

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