Question

If $f$ and $g$ are differentiable functions such that $f\left(2\right)=3,f\text{'}\left(2\right)=1,f\text{'}\left(3\right)=7,g\left(2\right)=-5$ and $g\text{'}\left(2\right)=2$, then find the numbers indicated in problem $\left(\frac{f}{f+g}\right)\text{'}\left(2\right)$.

Answer 1

To find the value of $\left(\frac{f}{f+g}\right)\text{'}\left(2\right)$.

If $f$ and $g$ are differentiable functions such that $f\left(2\right)=3,f\text{'}\left(2\right)=1,f\text{'}\left(3\right)=7,g\left(2\right)=-5$ and $g\text{'}\left(2\right)=2$.

Calculate the value of $\left(\frac{f}{f+g}\right)\text{'}\left(2\right)$ in below :

$\begin{array}{rcl}\left(\frac{f}{f+g}\right)\text{'}\left(2\right)& =& \frac{f\text{'}\left(2\right)}{\left(f+g\right)\text{'}\left(2\right)}\\ & \Rightarrow & \left(\frac{f}{f+g}\right)\text{'}\left(2\right)=\frac{1}{f\text{'}\left(2\right)+g\text{'}\left(2\right)}\left[\text{Substitute values}\right]\\ \therefore \left(\frac{f}{f+g}\right)\text{'}\left(2\right)& =& \frac{1}{1+2}=\frac{1}{3}\left[\text{Simplifying}\right]\end{array}$

Hence, the value of $\left(\frac{f}{f+g}\right)\text{'}\left(2\right)$ is $\frac{\mathbf{1}}{\mathbf{3}}$.