Q. 12 - Calculus

Question

Given that f,g,h are functions with values f(2)=1, g(2)=-4, h(2)=3 and point -derivatives f'(2)=3, g'(2)=0, h'(2)=-1 Calculate

$(2f+3g-h)\text{'}\left(2\right)$

Step-by-Step Solution

Answer

Hence the value is $(2f+3g-h)\text{'}\left(2\right)=7$

Step 1: Given information

We are given that f,g,h are functions with values f(2)=1, g(2)=-4, h(2)=3 and point -derivatives f'(2)=3, g'(2)=0, h'(2)=-1 

Step 2: Compute the derivative

We are given $(2f+3g-h)\text{'}\left(2\right)$

Which can be simplified as

$$\begin{gathered} 2f\text{'}\left(2\right)+3g\text{'}\left(2\right)-h\text{'}\left(2\right) \\ 2\left(3\right)+3\left(0\right)+1 \\ 6+1 \\ =7 \end{gathered}$$

hence $(2f+3g-h)\text{'}\left(2\right)=7$