Advertisements
Advertisements
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
Advertisements
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.
shaalaa.com
Is there an error in this question or solution?
