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Question Number 109119 by 1777 last updated on 21/Aug/20

$$iff\left(x^2\right)=y,f^{\prime }\left(x\right)=\sqrt{5x-1}then$$$$\frac{dy}{dx}=.....$$

Answered by bemath last updated on 21/Aug/20

$$\frac{△\mathrm{♭}e\mathcal{M}ath▽}{\bullet °----\bullet °}$$$$\Rightarrow f\left(x\right)=\int f{}^{\prime }\left(x\right)dx$$$$f\left(x\right)=\frac{1}{5}\int \sqrt{5x-1}d(5x-1)$$$$f\left(x\right)=\frac{2}{15}{(5x-1)}^{\frac{3}{2}}+C$$$$soy=f\left(x^2\right)=\frac{2}{15}{(5x^2-1)}^{\frac{3}{2}}+C$$$$\frac{dy}{dx}=\frac{2}{15}.\frac{3}{2}.10x\sqrt{5x^2-1}$$$$=2x\sqrt{5x^2-1}$$

Answered by 1549442205PVT last updated on 21/Aug/20

$$\mathrm{Set}\mathrm{u}={\mathrm{x}}^2\Rightarrow \mathrm{y}=\mathrm{f}\left(\mathrm{u}\right)\Rightarrow \frac{\mathrm{dy}}{\mathrm{du}}=\mathrm{f}{}^{\prime }\left(\mathrm{u}\right)=\sqrt{5\mathrm{u}-1}$$$$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy}}{\mathrm{du}}\times \frac{\mathrm{du}}{\mathrm{dx}}=\mathrm{f}{}^{\prime }\left(\mathrm{u}\right)\times {\mathrm{u}}^{\prime }\left(\mathrm{x}\right)$$$$=\sqrt{{5\mathrm{x}}^2-1}\times 2\mathrm{x}$$

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