Question

Using the limit definition how do you differentiate $f\left( x \right)=\dfrac{1}{\sqrt{x}}$

Answer 1

Hint: To find the differentiation of the function $f\left( x \right)=\dfrac{1}{\sqrt{x}}$ we will use the limit definition of the function which is given by $f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$ . Now we will substitute f(x) and take LCM of the obtained equation to simplify. We further multiply and divide by conjugate of the numerator and simplify the limit by substituting h = 0. Hence we get the differentiation of the required function.

Complete step-by-step solution:
Now limit definition of differentiation of any function f(x) is given as $f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$
Here we are given the function $f\left( x \right)=\dfrac{1}{\sqrt{x}}$
Hence substituting the function in the definition of limits we get,
$\Rightarrow f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{1}{\sqrt{x+h}}-\dfrac{1}{\sqrt{x}}}{h}$
Taking h in numerator we get,
$\Rightarrow f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{1}{h\sqrt{x+h}}-\dfrac{1}{h\sqrt{x}}$
Now taking LCM and making denominators common we get,
$\Rightarrow f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{h\sqrt{x}}{h\sqrt{x+h}}\dfrac{1}{h\sqrt{x}}-\dfrac{h\sqrt{x+h}}{h\sqrt{x}}\dfrac{1}{h\sqrt{x+h}}$
Now we know that $\dfrac{a}{b}-\dfrac{c}{b}=\dfrac{a-c}{b}$ Hence using this we get,
$\Rightarrow f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{h\sqrt{x}-h\sqrt{x+h}}{{{h}^{2}}\sqrt{x}\sqrt{x+h}}$
Now take h common from the numerator and cancel it with the h in the denominator. Hence, we get
$\Rightarrow f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sqrt{x}-\sqrt{x+h}}{h\sqrt{x}\sqrt{x+h}}$
Now multiplying the numerator and denominator of the above equation by conjugate of numerator which is $\sqrt{x}+\sqrt{x+h}$ we get,
$\Rightarrow f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sqrt{x}-\sqrt{x+h}}{h\sqrt{x}\sqrt{x+h}}\times \dfrac{\sqrt{x}+\sqrt{x+h}}{\sqrt{x}+\sqrt{x+h}}$
Now we know that $\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a-b \right)\left( a+b \right)$ . Hence using this we get,
 \[\begin{align}
  & \Rightarrow f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{x-x-h.}{h\left( \sqrt{x}+\sqrt{x+h} \right)\sqrt{x\left( x+h \right)}} \\
 & \Rightarrow f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{-1}{\left( \sqrt{x}+\sqrt{x+h} \right)\sqrt{x\left( x+h \right)}} \\
\end{align}\]
Now substituting the limits we get,
\[\begin{align}
  & \Rightarrow f'\left( x \right)=\dfrac{-1}{\left( \sqrt{x}+\sqrt{x} \right)\sqrt{x\left( x \right)}} \\
 & \Rightarrow f'\left( x \right)= -\dfrac{1}{2x\sqrt{x}}= -\dfrac{1}{2{{x}^{\dfrac{3}{2}}}} \\
\end{align}\]
Hence we get the derivative of the function $\dfrac{1}{\sqrt{x}}$ is $-\dfrac{1}{2{{x}^{\dfrac{3}{2}}}}$ .

Note: To understand the limits definition of a function we can imagine a graph of function. The points f(x) and f(x+h) correspond to the value of the function on the y axis corresponding to points x and x+h. Hence if we check the slope of the function it will be given by $\dfrac{f\left( x+h \right)-f\left( x \right)}{x+h-x}=\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$ . Now if h is very small the points are very close to each other and hence the function can be approximated as a straight line. Therefore we take $\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$ as differentiation of the function.