Question

find the derivative of (5x/cube root of 1-x square)+sin square(2x+3)

Answer 1

$$\begin{gathered} \frac{\mathrm{d}}{\mathrm{dx}}\left[\left(\frac{5\mathrm{x}}{\sqrt[3]{1-{\mathrm{x}}^2}}\right)+{\sin }^2\left(2\mathrm{x}+3\right)\right] \\ =\frac{\mathrm{d}}{\mathrm{dx}}\frac{5\mathrm{x}}{\sqrt[3]{1-{\mathrm{x}}^2}}+\frac{\mathrm{d}}{\mathrm{dx}}{\sin }^2\left(2\mathrm{x}+3\right) \\ =\frac{\sqrt[3]{1-{\mathrm{x}}^2}\frac{\mathrm{d}}{\mathrm{dx}}5\mathrm{x}-5\mathrm{x}\frac{\mathrm{d}}{\mathrm{dx}}{\left(1-{\mathrm{x}}^2\right)}^{{\displaystyle \frac{1}{3}}}}{{\left(\sqrt[3]{1-{\mathrm{x}}^2}\right)}^2}+2\sin \left(2\mathrm{x}+3\right)\frac{\mathrm{d}}{\mathrm{dx}}\sin \left(2\mathrm{x}+3\right) \\ =\frac{5\sqrt[3]{1-{\mathrm{x}}^2}-{\displaystyle \frac{5\mathrm{x}}{3}}{\left(1-{\mathrm{x}}^2\right)}^{{\displaystyle \frac{1}{3}-1}}{\displaystyle \frac{\mathrm{d}}{\mathrm{dx}}}\left(1-{\mathrm{x}}^2\right)}{{\left(\sqrt[3]{1-{\mathrm{x}}^2}\right)}^2}+2\sin \left(2\mathrm{x}+3\right)\cos \left(2\mathrm{x}+3\right)\frac{\mathrm{d}}{\mathrm{dx}}\left(2\mathrm{x}+3\right) \\ =\frac{5\sqrt[3]{1-{\mathrm{x}}^2}+\frac{10{\mathrm{x}}^2}{3}{\left(1-{\mathrm{x}}^2\right)}^{\frac{-2}{3}}}{{\left(\sqrt[3]{1-{\mathrm{x}}^2}\right)}^2}+2\sin 2\left(2\mathrm{x}+3\right) \\ =\frac{5\sqrt[3]{1-{\mathrm{x}}^2}}{{\left(\sqrt[3]{1-{\mathrm{x}}^2}\right)}^2}+\frac{10{\mathrm{x}}^2{\left(1-{\mathrm{x}}^2\right)}^{\frac{-2}{3}}}{3{\left(\sqrt[3]{1-{\mathrm{x}}^2}\right)}^2}+2\sin 2\left(2\mathrm{x}+3\right) \\ =\frac{5}{\sqrt[3]{1-{\mathrm{x}}^2}}+\frac{10{\mathrm{x}}^2}{3{\left(\sqrt[3]{1-{\mathrm{x}}^2}\right)}^{2+{\displaystyle \frac{2}{3}}}}+2\sin 2\left(2\mathrm{x}+3\right) \\ =\frac{5}{\sqrt[3]{1-{\mathrm{x}}^2}}+\frac{10{\mathrm{x}}^2}{3{\left(\sqrt[3]{1-{\mathrm{x}}^2}\right)}^{{\displaystyle \frac{8}{3}}}}+2\sin 2\left(2\mathrm{x}+3\right) \end{gathered}$$