Question

If $f\left(x\right)=\log \left\{\frac{u\left(x\right)}{v\left(x\right)}\right\},u\left(1\right)=v\left(1\right)\mathrm{and}u\text{'}\left(1\right)=v\text{'}\left(1\right)=2$, then find the value of f' (1).

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have},f\left(x\right)=\log \left\{\frac{u\left(x\right)}{v\left(x\right)}\right\} \\ \mathrm{and}, \\ u\left(1\right)=v\left(1\right),u\text{'}\left(1\right)=v\text{'}\left(1\right)=2...\left(i\right) \end{gathered}$$

$$\begin{gathered} \Rightarrow f\text{'}\left(x\right)=\frac{d}{dx}\left[\log \left\{\frac{u\left(x\right)}{v\left(x\right)}\right\}\right] \\ \Rightarrow f\text{'}\left(x\right)=\frac{1}{\left[{\displaystyle \frac{u\left(x\right)}{v\left(x\right)}}\right]}\times \frac{d}{dx}\left[\frac{u\left(x\right)}{v\left(x\right)}\right] \\ \Rightarrow f\text{'}\left(x\right)=\frac{v\left(x\right)}{u\left(x\right)}\times \left[\frac{v\left(x\right){\displaystyle \frac{d}{dx}}\left\{u\left(x\right)\right\}-u\left(x\right){\displaystyle \frac{d}{dx}\left\{v\left(x\right)\right\}}}{{\left\{v\left(x\right)\right\}}^2}\right] \\ \Rightarrow f\text{'}\left(x\right)=\frac{v\left(x\right)}{u\left(x\right)}\times \left[\frac{v\left(x\right)\times u\text{'}\left(x\right)-u\left(x\right)\times v\text{'}\left(x\right)}{{\left\{v\left(x\right)\right\}}^2}\right] \\ \mathrm{Putting}x=1,\mathrm{we}\mathrm{get}, \\ f\text{'}\left(1\right)=\frac{v\left(1\right)}{u\left(1\right)}\times \left[\frac{v\left(1\right)\times u\text{'}\left(1\right)-u\left(1\right)\times v\text{'}\left(1\right)}{{\left\{v\left(1\right)\right\}}^2}\right] \\ \Rightarrow f\text{'}\left(1\right)=1\times \left[\frac{u\left(1\right)\times 2-u\left(1\right)\times 2}{{\left\{u\left(1\right)\right\}}^2}\right]\left[\mathrm{Using}\mathrm{eqn}\left(1\right)\right] \\ \Rightarrow f\text{'}\left(1\right)=\left[\frac{0}{{\left\{u\left(1\right)\right\}}^2}\right] \\ \Rightarrow f\text{'}\left(1\right)=0 \end{gathered}$$