Question

How do you find the derivative of $y=\sqrt{x-3}$ using the limit definition?

Answer 1

In this problem, we have to find the derivative of the given expression using the limit definition. We should know the Limit definition to solve this problem. We have to substitute the given function in the limit definition formula to get the derivative of the given function.

Complete step by step answer:
We know that the given function to which we have to find the derivative is,
$f\left( x \right)=y=\sqrt{x-3}$…… (1)
We can write the above function as,
$f\left( x+h \right)=y=\sqrt{x+h-3}$….. (2)
We know that the limit definition is,
$f'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$ …… (3)
Now we can substitute the functions (1) and (2) in the limit definition (3), we get
$\Rightarrow f'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{\sqrt{x+h-3}-\sqrt{x-3}}{h}$
Now we can multiply the numerator and denominator with conjugate of numerator $\dfrac{\sqrt{x+h-3}+\sqrt{x-3}}{\sqrt{x+h-3}+\sqrt{x-3}}$, in order to cancel the radicals, we get
$\Rightarrow f'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{\sqrt{x+h-3}-\sqrt{x-3}}{h}\times \dfrac{\sqrt{x+h-3}+\sqrt{x-3}}{\sqrt{x+h-3}+\sqrt{x-3}}$
Now we can cancel the radical in the numerator, by the property $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$and we can multiply the denominator, we get

& \Rightarrow f'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{{{\left( \sqrt{x+h-3} \right)}^{2}}-{{\left( \sqrt{x-3} \right)}^{2}}}{h\left( \sqrt{x+h-3}+\sqrt{x-3} \right)} \\\ & \Rightarrow f'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{\left( x+h-3 \right)-\left( x-3 \right)}{h\left( \sqrt{x+h-3}+\sqrt{x-3} \right)} \\\ \end{aligned}$$ Now we can simplify the numerator in the above step, we get $$\Rightarrow f'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{h}{h\left( \sqrt{x+h-3}+\sqrt{x-3} \right)}$$ We can cancel the similar terms in the numerator and the denominator, we get $$\Rightarrow f'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{1}{\left( \sqrt{x+h-3}+\sqrt{x-3} \right)}$$ Now, we can apply the given limit value, we get $$\begin{aligned} & \Rightarrow f'\left( x \right)=\dfrac{1}{\left( \sqrt{x+0-3}+\sqrt{x-3} \right)} \\\ & \Rightarrow f'\left( x \right)=\dfrac{1}{\left( \sqrt{x-3}+\sqrt{x-3} \right)} \\\ \end{aligned}$$ We can now add the denominator, we get $$\Rightarrow f'\left( x \right)=\dfrac{1}{\left( 2\sqrt{x-3} \right)}$$ Therefore, the answer is $$f'\left( x \right)=\dfrac{1}{\left( 2\sqrt{x-3} \right)}$$. **Note:** Students make mistakes while simplifying the steps, we have more simplification in these types of problems. To solve these types of problems, we have to know the limit definition formula, in which we can apply the given function directly to find the derivative.