Question

How do you use the limit definition to find the derivative of $f\left( x \right)=\sqrt{x-7}$?

Answer 1

In order to find the solution of the given question that is to use the limit definition to find the derivative of $f\left( x \right)=\sqrt{x-7}$ apply the limit definition of the derivative that is ${f}'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$ and solve it further to find the derivative.

Complete step by step answer:
According to the question, given function in the question is as follows:
$f\left( x \right)=\sqrt{x-7}$
Now apply the limit definition of the derivative that is ${f}'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$ in the above expression, we will have:
$\Rightarrow {f}'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{\sqrt{\left( x+h \right)-7}-\sqrt{x-7}}{h}$
To simplify the expression from the right-hand side of the above equation, rationalise it by multiplying and dividing the term $\sqrt{\left( x+h \right)-7}+\sqrt{x-7}$, we will have:
$\Rightarrow {f}'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{\sqrt{\left( x+h \right)-7}-\sqrt{x-7}}{h}\cdot \dfrac{\sqrt{\left( x+h \right)-7}+\sqrt{x-7}}{\sqrt{\left( x+h \right)-7}+\sqrt{x-7}}$
Solve the numerator of the expression from the right-hand side of the above equation by using the formula $\left( a-b \right)\left( a+b \right)=\left( {{a}^{2}}-{{b}^{2}} \right)$, we will get:
$\Rightarrow {f}'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{\left( x+h-7 \right)-\left( x-7 \right)}{h\cdot \left( \sqrt{\left( x+h \right)-7}+\sqrt{x-7} \right)}$
Clearly, we can see that after opening the bracket some terms from the numerator of the expression from the right-hand side of the above equation are cancelling with each other, hence we are left with following:
$\Rightarrow {f}'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{h}{h\cdot \left( \sqrt{\left( x+h \right)-7}+\sqrt{x-7} \right)}$
As we can see that $h$ is there in both numerator and denominator of the expression in the above equation which means they will get cancelled and we will have:
$\Rightarrow {f}'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{1}{\sqrt{\left( x+h \right)-7}+\sqrt{x-7}}$
Now apply the limit $h \to 0$ in the right-hand side of the above equation, we will get:
$\Rightarrow {f}'\left( x \right)=\dfrac{1}{\sqrt{x-7}+\sqrt{x-7}}$
After simplifying the above equation, we will get the final answer as follows:
$\Rightarrow {f}'\left( x \right)=\dfrac{1}{2\sqrt{x-7}}$
Therefore, the derivative of $f\left( x \right)=\sqrt{x-7}$ is equal to $\dfrac{1}{2\sqrt{x-7}}$.

Note:
Students can go wrong by applying the wrong formula of limit definition of derivative that is they apply ${f}'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{f\left( x \right)-f\left( x+h \right)}{h}$which is completely wrong and leads to the wrong answer. It’s important to remember that the formula of limit definition of derivative is ${f}'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$.