Question

How do you find the derivative of $\sqrt {1 - 2x}$ ?

Answer 1

Hint : In order to determine the differentiation of the above function with respect to x, we will be using chaining rule by considering $1 - 2x$ as X and using derivative rule $\dfrac{d}{{dx}}(\sqrt X ) = \dfrac{1}{2}.{X^{^{ - \,\dfrac{1}{2}}}}\dfrac{d}{{dx}}(X)$ .The derivative of any constant term is zero and that of variable $(x)$ is equal to $1$ .Using these properties of derivative you will get your required answer.

Complete step by step solution:
We are Given a expression $\sqrt {1 - 2x}$ and we have to find the derivative of this expression with respect to x. Let this function be $f(x)$
$f(x) = \sqrt {1 - 2x}$
We have to find the first derivative of the above equation
$\dfrac{d}{{dx}}\left[ {f(x)} \right] = f'(x) \\\ f'(x) = \dfrac{d}{{dx}}\left[ {\sqrt {1 - 2x} } \right] \;$
Let’s Assume $1 - 2x$ as X
$f'(x) = \dfrac{d}{{dx}}\sqrt X$ -------(1)
Now Applying Chain rule to the above derivative which says that if we are not given a single variable $x$ and instead of it a function is given( $X$ )then the derivative will become
$\dfrac{d}{{dx}}(\sqrt X ) = \dfrac{d}{{dx}}({X^{\dfrac{1}{2}}}) = \dfrac{1}{2}.{X^{^{ - \,\dfrac{1}{2}}}}\dfrac{d}{{dx}}(X)$
We know that Derivative of variable $x$ raised to power some value $n$ is $\dfrac{d}{{dx}}({x^n}) = n{x^{n - 1}}$
Putting $\dfrac{d}{{dx}}(\sqrt X ) = \dfrac{1}{2}.{X^{^{ - \,\dfrac{1}{2}}}}\dfrac{d}{{dx}}(X)$ in the equation (1)
$f'(x) = \dfrac{1}{2}.{X^{^{ - \,\dfrac{1}{2}}}}\dfrac{d}{{dx}}(X)$
Putting back $X$
$f'(x) = \dfrac{1}{2}{\left( {1 - 2x} \right)^{^{ - \,\dfrac{1}{2}}}}\left( {\dfrac{d}{{dx}}(1 - 2x)} \right) \\\ f'(x) = \dfrac{1}{2}{\left( {1 - 2x} \right)^{^{ - \,\dfrac{1}{2}}}}\left( {\dfrac{d}{{dx}}(1) - 2\dfrac{d}{{dx}}(x)} \right) \;$
Since $\dfrac{d}{{dx}}(x) = 1$ , $\dfrac{d}{{dx}}(C) = 0$ ,where C is any constant number
$f'(x) = \dfrac{1}{2}{\left( {1 - 2x} \right)^{^{ - \,\dfrac{1}{2}}}}\left( {0 - 2(1)} \right) \\\ f'(x) = \dfrac{1}{2}{\left( {1 - 2x} \right)^{^{ - \,\dfrac{1}{2}}}}\left( { - 2} \right) \\\ f'(x) = - {\left( {1 - 2x} \right)^{^{ - \,\dfrac{1}{2}}}} \\\ f'(x) = - \dfrac{1}{{\sqrt {1 - 2x} }} \;$
Therefore, the derivative of $f(x) = \sqrt {1 - 2x}$ with respect to x is equal to $- \dfrac{1}{{\sqrt {1 - 2x} }}$
So, the correct answer is “ $- \dfrac{1}{{\sqrt {1 - 2x} }}$ ”.

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:
$\dfrac{{dy}}{{dx}}$ .
1.Use standard formula carefully while evaluating the derivative.
2. The $\dfrac{{dy}}{{dx}}$ is read as the derivative of $y$ with respect to x.
3. Chain rule helps to find the derivative of functions which are composite in nature.
4.Derivatives are used to find out the profit and loss in the business from the graph .