Q. 79 - Calculus

Question

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

Step-by-Step Solution

Answer

Proved

Step 1. To prove

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

Step 2. Proving

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