
What is the derivative of \[\sqrt{2x-1}\]?
Answer
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Hint: We are asked to find the derivative of \[\sqrt{2x-1}\]. In order to find the derivative, we are supposed to apply the chain rule. The general form that would be applied is \[\dfrac{dy}{dx}f\left( g\left( x \right) \right)=f'(g(x)).g'(x)\]. Firstly, we would be converting into the required form and then differentiating it gives us the required derivative of \[\sqrt{2x-1}\].
Complete step by step solution:
Now let us learn about the chain rule. Chain rule is a formula that is used to compute the derivative of a function. The chain rule equation is \[\dfrac{dy}{dx}f\left( g\left( x \right) \right)=f'(g(x)).g'(x)\]. This helps us to differentiate the composite functions. The chain rule can be reversed by integration by substitution method.
Now let us evaluate the derivative of \[\sqrt{2x-1}\]
By applying the chain rule i.e. \[\dfrac{dy}{dx}f\left( g\left( x \right) \right)=f'(g(x)).g'(x)\]
We can rewrite \[\sqrt{2x-1}\] as \[{{\left( 2x-1 \right)}^{\dfrac{1}{2}}}\]
On differentiating \[{{\left( 2x-1 \right)}^{\dfrac{1}{2}}}\] by applying \[\dfrac{dy}{dx}{{a}^{n}}=n\centerdot {{a}^{n-1}}\]
We get, \[\dfrac{1}{2}{{\left( 2x-1 \right)}^{-\dfrac{1}{2}}}\]
On equating this function to the chain rule, we get-
\[\dfrac{dy}{dx}f\left( g\left( x \right) \right)=\dfrac{1}{2}{{\left( 2x-1 \right)}^{-\dfrac{1}{2}}}\]
From the above equation, \[g\left( x \right)=2x-1\]
\[\Rightarrow g\grave{\ }\left( x \right)=2\]
Consider, \[f'(g(x)).g'(x)\]
\[\Rightarrow \]\[\dfrac{1}{2}{{\left( 2x-1 \right)}^{-\dfrac{1}{2}}}\centerdot 2\]
Upon cancelling \[2\], we get
\[\Rightarrow {{\left( 2x-1 \right)}^{-\dfrac{1}{2}}}\]
This can also be written as \[\dfrac{1}{\sqrt{2x-1}}\]
\[\therefore \] The derivative of \[\sqrt{2x-1}\] is \[\dfrac{1}{\sqrt{2x-1}}\].
Note: The chain rule is called chain rule because it is used to take derivatives of composite functions by chaining their derivatives together as we have done in the problem. There is a special case in chain rule that is power rule. it is useful when finding the derivative of a function that is raised to the \[{{n}^{th}}\]power . The most common error that happens is not being able to recognise the inner and outer function correctly. Computing the equation wrongly is another common error.
Complete step by step solution:
Now let us learn about the chain rule. Chain rule is a formula that is used to compute the derivative of a function. The chain rule equation is \[\dfrac{dy}{dx}f\left( g\left( x \right) \right)=f'(g(x)).g'(x)\]. This helps us to differentiate the composite functions. The chain rule can be reversed by integration by substitution method.
Now let us evaluate the derivative of \[\sqrt{2x-1}\]
By applying the chain rule i.e. \[\dfrac{dy}{dx}f\left( g\left( x \right) \right)=f'(g(x)).g'(x)\]
We can rewrite \[\sqrt{2x-1}\] as \[{{\left( 2x-1 \right)}^{\dfrac{1}{2}}}\]
On differentiating \[{{\left( 2x-1 \right)}^{\dfrac{1}{2}}}\] by applying \[\dfrac{dy}{dx}{{a}^{n}}=n\centerdot {{a}^{n-1}}\]
We get, \[\dfrac{1}{2}{{\left( 2x-1 \right)}^{-\dfrac{1}{2}}}\]
On equating this function to the chain rule, we get-
\[\dfrac{dy}{dx}f\left( g\left( x \right) \right)=\dfrac{1}{2}{{\left( 2x-1 \right)}^{-\dfrac{1}{2}}}\]
From the above equation, \[g\left( x \right)=2x-1\]
\[\Rightarrow g\grave{\ }\left( x \right)=2\]
Consider, \[f'(g(x)).g'(x)\]
\[\Rightarrow \]\[\dfrac{1}{2}{{\left( 2x-1 \right)}^{-\dfrac{1}{2}}}\centerdot 2\]
Upon cancelling \[2\], we get
\[\Rightarrow {{\left( 2x-1 \right)}^{-\dfrac{1}{2}}}\]
This can also be written as \[\dfrac{1}{\sqrt{2x-1}}\]
\[\therefore \] The derivative of \[\sqrt{2x-1}\] is \[\dfrac{1}{\sqrt{2x-1}}\].
Note: The chain rule is called chain rule because it is used to take derivatives of composite functions by chaining their derivatives together as we have done in the problem. There is a special case in chain rule that is power rule. it is useful when finding the derivative of a function that is raised to the \[{{n}^{th}}\]power . The most common error that happens is not being able to recognise the inner and outer function correctly. Computing the equation wrongly is another common error.
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