Question - Trigonometric and Logarithmic Equations

\text{Given the equations:}

[1]\: y = \sin^3{x} + \csc^5{x^3} + \tan^5{ \left( \sqrt{x^2 + 1^3} \right)}

\text{Differentiate equation [1] with respect to } x:

\frac{d}{dx}y = 3\sin^2{x}\cos{x} - 5\csc^6{x^3}(3x^2)\cot{x^3} + 5\tan^4{ \left( \sqrt{x^2 + 1^3} \right) }\frac{1}{2\sqrt{x^2 + 1^3}}(2x)(1+\tan^2{ \left( \sqrt{x^2 + 1^3} \right) })

[2]\: y = x\cos(\ln{x}) + 10\sec{(\sqrt{x})}

\text{Differentiate equation [2] with respect to } x:

\frac{d}{dx}y = \cos(\ln{x}) - x\sin(\ln{x})\frac{1}{x} + 10\sec{(\sqrt{x})}\tan{(\sqrt{x})}\frac{1}{2\sqrt{x}}

[3]\: y = \frac{\sin{(x^2-1)}}{x} + \cos{(\cos{x})^2} - 2\cot{(x^{-3})}

\text{Differentiate equation [3] with respect to } x:

\frac{d}{dx}y = \frac{x(2x)\cos{(x^2 - 1)} - \sin{(x^2 - 1)}}{x^2} - 2\cos{(\cos{x})}\sin{x}(-\sin{x}) + 2\csc^2{(x^{-3})}(3x^{-4})