Question - Calculate the Derivative of a Trigonometric Function

\frac{dy}{dx} = \frac{d}{dx}(sin^3(x) + \csc^5(x) + \tan^5(\sqrt{x^2 + 1}))

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

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

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