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\begin{align*} &\frac{d}{dx}\left[\sqrt{3x+4}\right]&&\text{Per the Chain rule:}\ \frac{df(u)}{dx}=\frac{df}{dx}\times \frac{du}{dx}\\ &=\frac{d}{du}\left[\sqrt{u}\right]\frac{d}{dx}\left[3x+4\right]&&\text{where}\ f=\sqrt{u},\ u=3x+4\\ &=\frac{1}{2\sqrt{u}}\frac{d}{dx}\left[3x+4\right]\\ &=\frac{1}{2\sqrt{3x+4}}\frac{d}{dx}\left[3x+4\right]&&\text{sub back in}\ u\\ &=\frac{1}{2\sqrt{3x+4}}\times 3\\ &=\frac{3}{2\sqrt{3x+4}}\\ \end{align*}
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Thu, 20 Jun 2024 12:45 GMT