Question

Solve this:

Q. Solve this :$\frac{d}{dx}\left(\cos x^3\right)=1$

Answer 1

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$$\begin{gathered} \frac{d}{dx}\left(\cos x^3\right)=1 \\ \left(-\sin x^3\right)\frac{d}{dx}\left(x^3\right)=0 \\ -3x^2\sin x^3=0 \\ {x}^2\sin x^3=0 \\ {x}^2=0or\sin x^3=0 \\ Consider, \\ {x}^2=0 \\ \Rightarrow x=0 \\ Nowconsider, \\ \sin x^3=0 \\ {x}^3=0+2n\mathrm{\pi },\mathrm{n}\in \mathrm{Z}\mathrm{or}{\mathrm{x}}^3=\mathrm{\pi }+2\mathrm{n\pi },\mathrm{n}\in \mathrm{Z} \\ {\mathrm{x}}^3=2\mathrm{n\pi },\mathrm{n}\in \mathrm{Z}\mathrm{or}{\mathrm{x}}^3=\mathrm{\pi }+2\mathrm{n\pi },\mathrm{n}\in \mathrm{Z} \\ \mathrm{x}=\sqrt[3]{2\mathrm{n\pi \pi }},\mathrm{n}\in \mathrm{Z}\mathrm{or}\mathrm{x}=\sqrt[3]{\mathrm{\pi }+2\mathrm{n\pi }},\mathrm{n}\in \mathrm{Z} \\ \mathrm{So},\mathrm{x}=0,\sqrt[3]{2\mathrm{n\pi \pi }},\mathrm{n}\in \mathrm{Z},\sqrt[3]{\mathrm{\pi }+2\mathrm{n\pi }},\mathrm{n}\in \mathrm{Z} \end{gathered}$$

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