
Differentiate ${{y}^{2}}=4ax$ with respect to $x$ (Where $a$ is a constant) ?
Answer
463.8k+ views
Hint: Here we have to differentiate the function given with respect to $x$ . As we can see that our independent variable has more than one power so we will use implicit differentiation. Firstly we will differentiate the function with respect to $x$ using the chain rule. Then we will collect all the $\dfrac{dy}{dx}$ terms on one side and simplify our value to get the desired answer.
Complete answer:
We have to differentiate the below function with respect to $x$ :
${{y}^{2}}=4ax$….$\left( 1 \right)$
Where $a$ is a constant
$\Rightarrow y=\sqrt{4ax}$…..$\left( 2 \right)$
We will use implicit differentiation in equation (1) as follows:
$\dfrac{d}{dx}\left( {{y}^{2}} \right)=\dfrac{d}{dx}\left( 4ax \right)$…$\left( 3 \right)$
Now as the Power rule and chain rule are given as follows:
Power Rule - $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$
Chain Rule - $\dfrac{d}{dx}{{y}^{n}}=n{{y}^{n-1}}\dfrac{dy}{dx}$
Also there will be no change in the constant $a$ as it is a coefficient of a variable.
Using the above rule in equation (3) we get,
$\Rightarrow 2{{y}^{2-1}}\dfrac{dy}{dx}=4a\times 1\times {{x}^{1-1}}$
$\Rightarrow 2y\dfrac{dy}{dx}=4a$
Keep the $\dfrac{dy}{dx}$ value on one term and take the rest term on another side as follows:
$\Rightarrow \dfrac{dy}{dx}=\dfrac{4a}{2y}$
On substituting the value from equation (2) above we get,
$\Rightarrow \dfrac{dy}{dx}=\dfrac{4a}{\sqrt{4ax}}$
$\Rightarrow \dfrac{dy}{dx}=\dfrac{4a}{2\sqrt{ax}}$
On simplifying we get,
$\Rightarrow \dfrac{dy}{dx}=\dfrac{2\sqrt{a}}{\sqrt{x}}$
$\Rightarrow \dfrac{dy}{dx}=2\sqrt{\dfrac{a}{x}}$
Hence derivative of ${{y}^{2}}=4ax$ with respect to $x$ (Where $a$ is a constant) is $2\sqrt{\dfrac{a}{x}}$ .
Note:
By using the chain rule we can differentiate an implicit function of $y$ with respect to $y$ but we have to multiply the result with $\dfrac{dy}{dx}$ as we have done in this question and this is known as implicit differentiation. Implicit differentiation is usually done when we are given a circle with center at origin. In this case we have both the variables with power as two and it becomes difficult to solve it in a simple manner. We have taken the constant as it is because it is in product with the variable but in case it is present individually then the differentiation of constant will be $0$ .
Complete answer:
We have to differentiate the below function with respect to $x$ :
${{y}^{2}}=4ax$….$\left( 1 \right)$
Where $a$ is a constant
$\Rightarrow y=\sqrt{4ax}$…..$\left( 2 \right)$
We will use implicit differentiation in equation (1) as follows:
$\dfrac{d}{dx}\left( {{y}^{2}} \right)=\dfrac{d}{dx}\left( 4ax \right)$…$\left( 3 \right)$
Now as the Power rule and chain rule are given as follows:
Power Rule - $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$
Chain Rule - $\dfrac{d}{dx}{{y}^{n}}=n{{y}^{n-1}}\dfrac{dy}{dx}$
Also there will be no change in the constant $a$ as it is a coefficient of a variable.
Using the above rule in equation (3) we get,
$\Rightarrow 2{{y}^{2-1}}\dfrac{dy}{dx}=4a\times 1\times {{x}^{1-1}}$
$\Rightarrow 2y\dfrac{dy}{dx}=4a$
Keep the $\dfrac{dy}{dx}$ value on one term and take the rest term on another side as follows:
$\Rightarrow \dfrac{dy}{dx}=\dfrac{4a}{2y}$
On substituting the value from equation (2) above we get,
$\Rightarrow \dfrac{dy}{dx}=\dfrac{4a}{\sqrt{4ax}}$
$\Rightarrow \dfrac{dy}{dx}=\dfrac{4a}{2\sqrt{ax}}$
On simplifying we get,
$\Rightarrow \dfrac{dy}{dx}=\dfrac{2\sqrt{a}}{\sqrt{x}}$
$\Rightarrow \dfrac{dy}{dx}=2\sqrt{\dfrac{a}{x}}$
Hence derivative of ${{y}^{2}}=4ax$ with respect to $x$ (Where $a$ is a constant) is $2\sqrt{\dfrac{a}{x}}$ .
Note:
By using the chain rule we can differentiate an implicit function of $y$ with respect to $y$ but we have to multiply the result with $\dfrac{dy}{dx}$ as we have done in this question and this is known as implicit differentiation. Implicit differentiation is usually done when we are given a circle with center at origin. In this case we have both the variables with power as two and it becomes difficult to solve it in a simple manner. We have taken the constant as it is because it is in product with the variable but in case it is present individually then the differentiation of constant will be $0$ .
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