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Question Number 211502 by MathematicalUser2357 last updated on 11/Sep/24

$$\mathrm{If}\{\begin{array}{l}f\left(x\right)=x^2\\ g\left(x\right)=\sin x\end{array},$$$$\mathrm{Then}\mathrm{find}\frac{df}{dg}.$$

Answered by a.lgnaoui last updated on 11/Sep/24

$$\frac{\mathrm{df}}{\mathrm{dg}}=\frac{\mathrm{df}}{\mathrm{dx}}\times \frac{\mathrm{dx}}{\mathrm{dg}}.=\frac{{\mathrm{f}}^{{}^{\prime }}}{{\mathrm{g}}^{\prime }}$$$${\mathrm{f}}^{\prime }=2\mathrm{x}{\mathrm{g}}^{\prime }=\cos \mathrm{x}$$$$\Rightarrow \frac{\mathrm{df}}{\mathrm{dg}}=\frac{2\mathbf{x}}{\cos \mathbf{x}}$$

Commented by MathematicalUser2357 last updated on 12/Sep/24

$$\mathrm{Can}\mathrm{be}\mathrm{solved}\mathrm{in}\mathrm{other}\mathrm{way}.$$$$\frac{df}{dg}=\frac{2xdx}{\cos xdx}=\frac{2x}{\cos x}=2x\sec x\left(\mathrm{No}\mathrm{need}\mathrm{to}\mathrm{transform}\frac{1}{\cos x}\mathrm{to}\sec x\right)$$

Answered by MATHEMATICSAM last updated on 11/Sep/24

$$f\left(x\right)=x^2\Rightarrow \frac{df}{dx}=2x$$$$g\left(x\right)=\sin x\Rightarrow \frac{dg}{dx}=\cos x$$$$\frac{df}{dg}=\frac{\frac{df}{dx}}{\frac{dg}{dx}}=\frac{2x}{\cos x}=2x\sec x$$

Commented by MathematicalUser2357 last updated on 12/Sep/24

$$$$🙏🙏🙏

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