Question

Show that if a function $y=f\left(x\right)$is differentiable at $x_0$and

$∆y=f(x_0+∆x)-f\left(x_0\right)$,

then

$∆y=f\text{'}\left(x_0\right)∆x+\in ∆x$,

where $\in$is a function satisfying $\underset{∆x\to 0}{\lim }\in =0$.

Short answer

Using the definition of derivatives to find the derivatives of the function

we have proved that,

if function $y=f\left(x\right)$is differentiable at $X_o$

$\mathrm{\Delta }y=f\left(x_0+\mathrm{\Delta }x\right)-f\left(x_0\right)$then

$\mathrm{\Delta }y=f^{\mathrm{\prime }}\left(x_0\right)\mathrm{\Delta }x+\in \mathrm{\Delta }x$

Step-by-step solution

Step1. Given Information

Consider a function

$f\left(x\right)$

$\text{Here the objective is to prove that}\mathrm{\Delta }y=f^{\mathrm{\prime }}\left(x_0\right)\mathrm{\Delta }x+\in \mathrm{\Delta }x.$

Step2. Derivative of the function

Use the definition of derivative to find the derivative of the function$x=x_0$

localid="1649762082441" $\begin{array}{l}\underset{\mathrm{\Delta }x\to 0}{lim} \frac{f\left(x_0+\mathrm{\Delta }x\right)-f\left(x_0\right)}{\mathrm{\Delta }x}=f^{\mathrm{\prime }}\left(x_0\right)\\ \frac{\underset{\mathrm{\Delta }x\to 0}{lim} f\left(x_0+\mathrm{\Delta }x\right)-f^{\mathrm{\prime }}\left(x_0\right)}{\underset{\mathrm{\Delta }x\to 0}{lim} \mathrm{\Delta }x}=f^{\mathrm{\prime }}\left(x_0\right)\\ \underset{\mathrm{\Delta }x\to 0}{lim} f\left(x_0+\mathrm{\Delta }x\right)-f\left(x_0\right)=f^{\mathrm{\prime }}\left(x_0\right)\cdot \underset{\mathrm{\Delta }x\to 0}{lim} \mathrm{\Delta }x\text{Use cr}\\ \underset{\mathrm{\Delta }x\to 0}{lim} \left[\mathrm{\Delta }y+f\left(x_0\right)\right]-\underset{\mathrm{\Delta }x\to 0}{lim} f\left(x_0\right)=f^{\mathrm{\prime }}\left(x_0\right)\mathrm{\Delta }x\\ \underset{\mathrm{\Delta }x\to 0}{lim} \mathrm{\Delta }y+\underset{\mathrm{\Delta }x\to 0}{lim} f\left(x_0\right)-\underset{\mathrm{\Delta }x\to 0}{lim} f\left(x_0\right)=f^{\mathrm{\prime }}\left(x_0\right)\mathrm{\Delta }x\\ \underset{\mathrm{\Delta }x\to 0}{lim} \mathrm{\Delta }y=f^{\mathrm{\prime }}\left(x_0\right)\mathrm{\Delta }x\\ \mathrm{\Delta }y=f^{\mathrm{\prime }}\left(x_0\right)\mathrm{\Delta }x+\in \mathrm{\Delta }x\end{array}$

the expression$\in \mathrm{\Delta }x$

has been added so that the equality holds such that

$\underset{\mathrm{\Delta }x\to 0}{lim} \in =0$