Question

How do you find the derivative of y = $\sqrt{5x}$

Answer 1

To find the derivative of the above mentioned question i.e. y = $\sqrt{5x}$ we will just have to substitute the derivative of x which we can get from differentiation of ${{x}^{n}}$ which is equal to $n{{x}^{n-1}}$.

Complete step by step solution:
Let us assume y = f(x) (as y is defined to be a function of x)
We have to find $\dfrac{df}{dx}$ which is none other than y’.
We can also write y’ =$\dfrac{df}{dx}$.
We already know how to calculate the differentiation of find i.e. $\dfrac{dy}{dx}$ = $n{{x}^{n-1}}$
Now by using the above stated property (or formula) we can easily calculate the value of$\sqrt{x}$.
Now the derivative of $\sqrt{x}$ is
⇒$\dfrac{dy}{dx}$ = $\dfrac{d{{x}^{1/2}}}{dx}$= $\dfrac{1}{2}{{x}^{\dfrac{1}{2}-1}}$= $\dfrac{1}{2}{{x}^{-\dfrac{1}{2}}}$ = $\dfrac{1}{2\sqrt{x}}$
So as we can see form the above statement or equation the differentiated value of $\sqrt{x}$ is equal to $\dfrac{1}{2\sqrt{x}}$ .
Now we will substitute this derivative of $\sqrt{x}$ in our main given function so as to finally get the differentiated value of the desired function:
We can conclude that $\dfrac{dy}{dx}$ =$\dfrac{d\sqrt{5x}}{dx}$
We can also write the above given statement as
$\dfrac{dy}{dx}$ = $\sqrt{5}\dfrac{d\sqrt{x}}{dx}$
We already have calculated the value of $\dfrac{d\sqrt{x}}{dx}$ which is equal to$\dfrac{1}{2\sqrt{x}}$ .
As we can see that $\sqrt{5}$ is not dependent on function x so we can say that $\sqrt{5}$ is a constant.
We also know that differentiation of a constant is equal to zero (0).
Now we can apply the multiplication rule (property) of differentiation which clearly states that:
If a function f(x) consists of two consecutive functions i.e. f(x) = g(x) h(x)
The derivative of that function will be equal to f’(x) = g’(x) h(x) + g(x) h’(x) ………….. (1)
Now on comparing the function f(x) with the question that is stated above we can say that
g(x) = $\sqrt{5}$ which is a constant
⇒g’(x) = 0 (which we have already calculated before)
h(x) = $\sqrt{x}$
⇒h’(x) = $\dfrac{1}{2\sqrt{x}}$ (which we have already calculated before)
Now with the help of multiplication rule and also by substituting all the values which we have calculated all in equation (1), we can finally get
⇒f’(x) = (0)( x) + ($\sqrt{5}$)( $\dfrac{1}{2\sqrt{x}}$)
⇒f’(x) = $\dfrac{\sqrt{5}}{2\sqrt{x}}$

So, from the above equation we can clearly say that the derivative of y = $\sqrt{5x}$ is $\dfrac{\sqrt{5}}{2\sqrt{x}}$

Note:
In this type of equation when there is a constant present as coefficient of the variable for which differentiation is taking place take the constant common and differentiate the variable like we did it in this question, we had $\sqrt{5}$ as a constant and $\sqrt{x}$ as the differentiating factor, we took the constant common (i.e. not performing any task on it) and only differentiating the variable which resulted us with the required answer.