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) \( x y+e^{x y}=5 \)

\[ \textbf{4(a)}\quad y = e^{x}\sin x. \] 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 \( y' \): \[ xy'=-y \quad \Longrightarrow \quad y'=-\frac{y}{x}. \]