Find the derivative of the following functions using logarithmic differentiation: (a) \(a(t)=\left(1+t^{2}\right)^{3}\left(1-t^{2}\right)^{4} \sin ^{3} t\) (b) \(b(r)=\mathrm{e}^{-r} \sin ^{5} 2 r \cos ^{4} 3 r\) (c) \(K(p)=6 p \sqrt{\sin p}(\cos 2 p)^{1 / 3}\) (d) \(N(y)=\frac{(1+\sqrt{y})^{4}(6+7 y)^{3}}{\sqrt{1+y}}\) (e) \(x(t)=\frac{6}{(2+t)^{3}(1-t)^{4} \cos ^{3} t}\)

Question: Find the derivatives of the following functions using logarithmic differentiation: (a) \(a(t)=\left(1+t^{2}\right)^{3}\left(1-t^{2}\right)^{4} \sin ^{3} t\) (b) \(b(r)=\mathrm{e}^{-r} \sin ^{5} 2 r \cos ^{4} 3 r\) (c) \(K(p)=6 p \sqrt{\sin p}(\cos 2 p)^{1 / 3}\) (d) \(N(y)=\frac{(1+\sqrt{y})^{4}(6+7 y)^{3}}{\sqrt{1+y}}\) (e) \(x(t)=\frac{6}{(2+t)^{3}(1-t)^{4} \cos ^{3} t}\) Answer: (a) \(a'(t) = (1+t^2)^3(1-t^2)^4\sin^3t\left( \frac{6t}{1+t^2} - \frac{8t}{1-t^2} + \frac{3\cos t}{\sin t} \right)\) (b) \(b'(r) = \mathrm{e}^{-r} \sin ^{5} 2 r \cos ^{4} 3 r \left( -1 + 5\frac{2\cos2r}{\sin2r} + 4\frac{-3\sin3r}{\cos3r} \right)\) (c) \(K'(p) = 6 p \sqrt{\sin p}(\cos 2 p)^{1 / 3}\left( \frac{1}{p} + \frac{1}{2}\frac{\cos p}{\sin p} - \frac{2}{3}\frac{\sin2p}{\cos2p} \right)\) (d) \(N'(y) = \frac{(1+\sqrt{y})^{4}(6+7 y)^{3}}{\sqrt{1+y}}\left( 4\frac{\frac{1}{2\sqrt{y}}}{1+\sqrt{y}} + 3\frac{7}{6+7y} - \frac{1}{2}\frac{1}{1+y} \right)\) (e) \(x'(t) = \frac{6}{(2+t)^{3}(1-t)^{4} \cos ^{3} t}\left( -\frac{3}{2+t} + \frac{4}{1-t} - \frac{3\sin t}{\cos t} \right)\)