Question

$$\begin{gathered} Th\mathrm{e}\mathrm{functions}\mathrm{F},\mathrm{G}\mathrm{are}\mathrm{differentiable}\mathrm{also} \\ G\left(x\right)=F\left(x\right)+C.\mathrm{Where},\mathrm{Cis}\mathrm{any}\mathrm{constant}. \\ P\mathrm{rove}\mathrm{that}F\text{'}\left(x\right)=G\left(x\right). \end{gathered}$$

Short answer

Proved

Step-by-step solution

Step 1. To prove

$F\text{'}\left(x\right)=G\left(x\right).$

Step 2. Proving

$$\begin{gathered} Now, \\ \frac{d}{dx}\left(G\right(x\left)\right)=\frac{d}{dx}\left(F\right(x)+C) \\ G\text{'}\left(x\right)=F\text{'}\left(x\right)+0 \\ G\text{'}\left(x\right)=F\text{'}\left(x\right)\mathrm{Therefore},\mathrm{it}\mathrm{is}\mathrm{proved}\mathrm{that}F\text{'}\left(x\right)=G\text{'}\left(x\right) \end{gathered}$$