Find \(\frac{\mathrm{d} y}{\mathrm{~d} x}\) where \(y\) is given by (a) \(x^{3} \mathrm{e}^{2 x}\) (b) \(-3 \sin 2 x \cos 5 x\) (c) \(\left(x^{2}+1\right) \mathrm{e}^{-x}\) (d) \(\sin x \sin 2 x \sin 3 x\) (e) \(x \tan 3 x\).

To find the derivative of \(x^3 e^{2x}\), we will use the product rule, which states that the derivative of the product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function. In this case, the two functions are \(u(x) = x^3\) and \(v(x) = e^{2x}\). We have: \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}u}{\mathrm{d}x}v + u\frac{\mathrm{d}v}{\mathrm{d}x}\) First, find the derivatives of u and v: \(\frac{\mathrm{d}u}{\mathrm{d}x} = 3x^2\) \(\frac{\mathrm{d}v}{\mathrm{d}x} = 2e^{2x}\) Now, find the derivative of y: \(\frac{\mathrm{d}y}{\mathrm{d}x} = (3x^2)(e^{2x}) + (x^3)(2e^{2x}) = 3x^2e^{2x} + 2x^3e^{2x} = (3x^2+2x^3)e^{2x}\)

To find the derivative of \(-3\sin(2x)\cos(5x)\), we will use the product rule again since we have a product of two functions \(u(x) = -3\sin(2x)\) and \(v(x) = \cos(5x)\). First, find the derivatives of u and v using the chain rule: \(\frac{\mathrm{d}u}{\mathrm{d}x} = -3(2\cos(2x)) = -6\cos(2x)\) \(\frac{\mathrm{d}v}{\mathrm{d}x} = -5\sin(5x)\) Now, find the derivative of y: \(\frac{\mathrm{d}y}{\mathrm{d}x} = (-6\cos(2x))(\cos(5x)) + (-3\sin(2x))(-5\sin(5x)) = -6\cos(2x)\cos(5x) + 15\sin(2x)\sin(5x)\)

To find the derivative of \((x^2+1)e^{-x}\), we will use the product rule again with the functions \(u(x) = x^2+1\) and \(v(x) = e^{-x}\). First, find the derivatives of u and v: \(\frac{\mathrm{d}u}{\mathrm{d}x} = 2x\) \(\frac{\mathrm{d}v}{\mathrm{d}x} = -e^{-x}\) Now, find the derivative of y: \(\frac{\mathrm{d}y}{\mathrm{d}x} = (2x)(e^{-x})+(x^2+1)(-e^{-x}) = 2xe^{-x} - (x^2+1)e^{-x}\)

To find the derivative, we will need to apply the product rule multiple times. First, consider the product \(\sin(x)\sin(2x)\) as a single function \(w(x)\). Then, apply the product rule to \(w(x)\sin(3x)\). Finally, we apply the product rule again to find the derivative of \(w(x)\) in terms of \(\sin(x)\) and \(\sin(2x)\). Let \(w(x) = \sin(x)\sin(2x)\). The derivative of \(w(x)\) is: \(\frac{\mathrm{d}w}{\mathrm{d}x} = (\cos(x))\sin(2x) + \sin(x)(2\cos(2x)) = \cos(x)\sin(2x)+2\sin(x)\cos(2x)\) Now, to find the derivative of the original function, use the product rule for \(w(x)\sin(3x)\): \(\frac{\mathrm{d}y}{\mathrm{d}x} = (\cos(x)\sin(2x)+2\sin(x)\cos(2x))(3\cos(3x))+ (\sin(x)\sin(2x))\sin(3x) = (\cos(x)\sin(2x)+2\sin(x)\cos(2x))3\cos(3x) + \sin(x)\sin(2x)\sin(3x)\)

To find the derivative of \(x\tan(3x)\), we will use the product rule with the functions \(u(x) = x\) and \(v(x) = \tan(3x)\). First, find the derivatives of u and v using the chain rule: \(\frac{\mathrm{d}u}{\mathrm{d}x} = 1\) \(\frac{\mathrm{d}v}{\mathrm{d}x} = (3\sec^2(3x))\) Now, find the derivative of y: \(\frac{\mathrm{d}y}{\mathrm{d}x} = (1)(\tan(3x)) + (x)(3\sec^2(3x)) = \tan(3x) + 3x\sec^2(3x)\)