Question

If $f\left(1\right)=3,f\text{'}\left(1\right)=2$, then $\frac{d}{dx}(\log f(e^x+2x\left)\right)atx=0$ is equal to

Answer 1

Finding the value of :$\frac{d}{dx}(\log f(e^x+2x\left)\right)atx=0$

Given that $f\left(1\right)=3,f\text{'}\left(1\right)=2$

$$\begin{gathered} \frac{d}{dx}(\log f(e^x+2x\left)\right)=\left[\frac{1}{f(e^x+2x)}\right]\times f\text{'}(e^x+2x)(e^x+2) \\ \Rightarrow =(e^x+2)\frac{f\text{'}(e^x+2x)}{f(e^x+2x)} \end{gathered}$$

Put $x=0$

$$\begin{gathered} \frac{d}{dx}(\log f{(e^x+2x))}_{x=0}=(e^0+2)\frac{f\text{'}(e^0+0)}{f(e^0+0)} \\ \Rightarrow =3\frac{f\text{'}\left(1\right)}{f\left(1\right)} \\ \Rightarrow =3\times \frac{2}{3} \\ \Rightarrow =2 \end{gathered}$$

Hence, The value of $\frac{d}{dx}(\log f(e^x+2x\left)\right)atx=0$ is equal to $2$