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After differentiating \[\log y=v(x)\cdot\log[u(x)]\], which equation is obtained?

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Question

After differentiating \[\log y=v(x)\cdot\log[u(x)]\], which equation is obtained?

Options

  • \[\frac{1}{y}\cdot\frac{dy}{dx}=v(x)\cdot\frac{d}{dx}(\log[u(x)])+\log[u(x)]\cdot\frac{d}{dx}(v(x))\]

  • \[y\cdot\frac{dy}{dx}=v(x)\cdot\frac{d}{dx}(\log[u(x)])\]

  • \[\frac{1}{y}\cdot\frac{dy}{dx}=\frac{d}{dx}(v(x))+\frac{d}{dx}(\log[u(x)])\]

  • \[\frac{dy}{dx}=v(x)\cdot\log[u(x)]\]

MCQ
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Solution

The chain rule gives \[\frac{1}{y}\frac{dy}{dx}\] for \[\log y\]. The product rule differentiates \[v(x)\cdot\log[u(x)]\] into the two terms shown.

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