Question

How do you find the derivative of $ \dfrac{1}{{\sqrt x }} $ ?

Answer 1

Hint: In order to find the first derivative of the above expression with respect to x Use the reciprocal rule of derivation i.e. $ {\left[ {\dfrac{1}{{u(x)}}} \right]^\prime } = \dfrac{{u'(x)}}{{u{{(x)}^2}}} $ ,considering $ u\left( x \right) = 1 - x $ to solve the above problem . . Later use the formula of differentiation that is Chain rule to solve further and simplify the result . The reason we are using the chain rule as this function can be written as a composition of two functions . The chain rule states $ \dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \times \dfrac{{du}}{{dx}} $ This means we have to differentiate both functions and multiply them and we can obtain the required answer .

Complete step-by-step answer:
Given a function $ \dfrac{1}{{\sqrt x }} $ let it be $ f(x) $
 $ f(x) = \dfrac{1}{{\sqrt x }} $
 $ y = \dfrac{1}{u} $ and $ u = \sqrt x = {x^{\dfrac{1}{2}}} $
On simplification further , we have that $ y = u $ and $ u = {x^{ - \dfrac{1}{2}}} $
Now we will apply the chain rule according to which $ \dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \times \dfrac{{du}}{{dx}} $ . So we have to differentiate both functions and multiply them . Starting with y –
By the power rule $ y' = 1 \times {u^0} = 1 $ . We have to find the first derivative of the above equation considering it as u : Again by the power rule we get : -
 $
  u' = - \dfrac{1}{2} \times {x^{ - \dfrac{1}{2} - 1}} \\
\Rightarrow u' = - \dfrac{1}{2}{x^{ - \dfrac{3}{2}}} \\
\Rightarrow u' = - \dfrac{1}{{2\sqrt {{x^3}} }} \\
\Rightarrow f'(x) = y' \times u' \\
\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \times \dfrac{{du}}{{dx}} \\
\Rightarrow f'(x) = 1 \times - \dfrac{1}{{2\sqrt {{x^3}} }} \\
\Rightarrow f'(x) = - \dfrac{1}{{2\sqrt {{x^3}} }} \;
  $
This $ f'(x) = - \dfrac{1}{{2\sqrt {{x^3}} }} $ is our required answer .
So, the correct answer is $ f'(x) = - \dfrac{1}{{2\sqrt {{x^3}} }} $.

Note: 1. Calculus consists of two important concepts one is differentiation and other is integration.
2.What is Differentiation?
It is a method by which we can find the derivative of the function .It is a process through which we can find the instantaneous rate of change in a function based on one of its variables.
Let y = f(x) be a function of x. So the rate of change of $ y $ per unit change in $ x $ is given by:
 $ \dfrac{{dy}}{{dx}} $.
3. Indefinite integral=Let $ f(x) $ be a function .Then the family of all its primitives (or antiderivatives) is called the indefinite integral of $ f(x) $ and is denoted by $ \int {f(x)} dx $
The symbol $ \int {f(x)dx} $ is read as the indefinite integral of $ f(x) $ with respect to x.