Q.27 - Calculus

Question

Suppose g(x), h(x), and j(x) are differentiable functions with

values of the function and its derivative given in the following table:

$if f\left(x\right)=\frac{3h\left(x\right)}{g\left(x\right) + j\left(x\right)}, find f \text{'}\left(0\right).$

Step-by-Step Solution

Answer

 The value of  $f\text{'}\left(0\right)$ is 162

Step1. Given information

Values of the function and its derivative given.

We have to find out the values of $f\text{'}\left(0\right)$

Step 2. Differentiate using the quotient rule

$$\begin{gathered} \mathrm{The} \mathrm{quotient} \mathrm{rule}:\mathrm{If} \mathrm{f} \mathrm{and} \mathrm{g} \mathrm{are} \mathrm{functions} \mathrm{and} \mathrm{both} \mathrm{f} \mathrm{and} \mathrm{g} \mathrm{are} \mathrm{differentiable}, \\ \mathrm{then} \mathrm{quotient} \mathrm{rule} {\left(\frac{\mathrm{f}}{\mathrm{g}}\right)}^{\text{'}}\left(\mathrm{x}\right)=\frac{\mathrm{f} \left(\mathrm{x}\right) \text{'}\mathrm{g}\left(\mathrm{x}\right) - \mathrm{f}\left(\mathrm{x}\right)\mathrm{g} \left(\mathrm{x}\right)\text{'}}{(\mathrm{g}{\left(\mathrm{x}\right))}^2} \\ \mathrm{Let} \mathrm{f}\left(\mathrm{x}\right)=3\mathrm{h}\left(\mathrm{x}\right) \\ \mathrm{g}\left(\mathrm{x}\right)=\mathrm{g}\left(\mathrm{x}\right)+\mathrm{j}\left(\mathrm{x}\right) \\ \mathrm{when} \mathrm{we} \mathrm{apply} \mathrm{quotient} \mathrm{rule} \mathrm{then} \mathrm{fiven} \mathrm{function} \\ \mathrm{f}\text{'}\left(\mathrm{x}\right)=\frac{3\mathrm{h}\text{'}\left(\mathrm{x}\right)\left(\mathrm{g}\right(\mathrm{x})+\mathrm{j}(\mathrm{x}\left)\right)-3\mathrm{h}\left(\mathrm{x}\right)(\mathrm{g}\text{'}(\mathrm{x})+\mathrm{j}\text{'}(\mathrm{x}\left)\right)}{{\left(\mathrm{g}\left(\mathrm{x}\right)+\mathrm{j}\left(\mathrm{x}\right)\right)}^2} \\ \mathrm{then},\mathrm{f}\text{'}\left(0\right)=\frac{3\mathrm{h}\text{'}\left(0\right)\left(\mathrm{g}\right(0)+\mathrm{j}(0\left)\right)-3\mathrm{h}\left(0\right)(\mathrm{g}\text{'}(0)+\mathrm{j}\text{'}(0\left)\right)}{{\left(\mathrm{g}\left(0\right)+\mathrm{j}\left(0\right)\right)}^2 } \\ \mathrm{from} \mathrm{the} \mathrm{given} \mathrm{table} \mathrm{we} \mathrm{get} \mathrm{values} \mathrm{of} \mathrm{h}\text{'}\left(0\right)=3,\mathrm{g}\left(0\right)=2,\mathrm{j}\left(0\right)=0 , \mathrm{g}\text{'}\left(0\right)=-2.\mathrm{j}\text{'}\left(0\right)=-2 \mathrm{and} \mathrm{h}\left(0\right)=3 \\ \mathrm{Substitute} \mathrm{these} \mathrm{values} \mathrm{in} \mathrm{f}\text{'}\left(0\right), \mathrm{then} \mathrm{we} \mathrm{get} \\ \mathrm{f}\text{'}\left(0\right)=\frac{3\times 3(2+0)-3\times 3(-2-2)}{{(2+0)}^2} \\ =\frac{18+36}{4} \\ =162 \end{gathered}$$