Question

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

Answer 1

Hint: Start by considering $ f(x) $ as the function. The next step is the substitution. Substitute the values in place of the terms to make the equation easier to solve. Then we apply the chain rule to solve the derivative. We will solve the derivatives of each term separately.

Complete step by step answer: First we will start off by considering $ \sqrt x = p $ so that the equation becomes $ y = \dfrac{1}{p} $ and hence $ u = {p^{\dfrac{1}{2}}} $ since $ \sqrt x = {x^{\dfrac{1}{2}}} $ .
Now if we simplify further we have $ y = p $ and $ p = {x^{\dfrac{{ - 1}}{2}}} $ .
Here we will apply the chain rule which is stated by:
$\Rightarrow$ $ \dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \times \dfrac{{du}}{{dx}} $
In our case, the equation will be:
$\Rightarrow$ $ \dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{dp}} \times \dfrac{{dp}}{{dx}} $
We will now have to differentiate both functions and then multiply them.
Here, we will start with $ y $.
$\Rightarrow$ So, by the power rule, $ y' = 1 \times {p^0} = 1 $ .
$\Rightarrow$ Now we will solve for $ p $.
 $
  p' = \dfrac{{ - 1}}{2} \times {x^{\dfrac{{ - 1}}{2} - 1}} \\
  p' = \dfrac{{ - 1}}{2} \times {x^{\dfrac{{ - 3}}{2}}} \\
  p' = \dfrac{{ - 1}}{{2\sqrt {{x^3}} }} \\
  $
$\Rightarrow$ As we know, $ f(x) = y \times p $
 $
  f'(x) = y' \times p' \\
  f'(x) = 1 \times - \dfrac{1}{{2\sqrt {{x^3}} }} \\
  f'(x) = - \dfrac{1}{{2\sqrt {{x^3}} }} \\
  $
$\Rightarrow$ Hence, the derivative of $ y = \dfrac{1}{{\sqrt x }} $ is $ - \dfrac{1}{{2\sqrt {{x^3}} }} $ .
Additional Information: A derivative is the rate of change of a function with respect to a variable. Derivatives are fundamental to the solution of problems in calculus and differential equations. In general, scientists observe changing systems to obtain the rate of change of some variable of interest, incorporate this information into some differential equation, and use integration techniques to obtain a function that can be used to predict the behavior of the original system under diverse conditions.

Note:
While substituting the terms make sure you are taking into account the degrees and signs of the terms as well. While applying the power rule make sure you have considered the power with their respective signs. Remember that the derivative of $ \sqrt x $ is $ \dfrac{1}{{2\sqrt x }} $ and the derivative of $ y $ is $ 1 $ .