
How do you differentiate the given function $\arcsin \left( 2x \right)$?
Answer
558.9k+ views
Hint: We start solving the problem by assuming \[2x=z\] and then differentiating both sides of the given function with respect to x. We then recall the chain rule of differentiation as $\dfrac{d\left( g\left( f \right) \right)}{dx}=\dfrac{d\left( g \right)}{df}\times \dfrac{df}{dx}$ to proceed through the problem. We then make use of the fact that $\dfrac{d\left( \arcsin \left( x \right) \right)}{dx}=\dfrac{1}{\sqrt{1-{{x}^{2}}}}$ to proceed through the problem. We then make use of the facts that $\dfrac{d\left( ax \right)}{dx}=a$ to get the required answer for the derivative of the function.
Complete step-by-step answer:
According to the problem, we are asked to find the derivative of the function $\arcsin \left( 2x \right)$.
Let us assume $y=\arcsin \left( 2x \right)$ ---(1).
Let us assume \[2x=z\]. Let us substitute this in equation (1).
$\Rightarrow y=\arcsin \left( z \right)$ ---(2).
Let us differentiate both sides of the equation (2) with respect to x.
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d\left( \arcsin \left( z \right) \right)}{dx}$ ---(3).
From chain rule of differentiation, we know that $\dfrac{d\left( g\left( f \right) \right)}{dx}=\dfrac{d\left( g \right)}{df}\times \dfrac{df}{dx}$. Let us substitute this result in equation (3).
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d\left( \arcsin \left( z \right) \right)}{dz}\times \dfrac{dz}{dx}$ ---(4).
We know that $\dfrac{d\left( \arcsin \left( x \right) \right)}{dx}=\dfrac{1}{\sqrt{1-{{x}^{2}}}}$. Let us use this result in equation (4).
$\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\sqrt{1-{{z}^{2}}}}\times \dfrac{dz}{dx}$ ---(5).
Now, let us substitute $z=2x$ in equation (5).
$\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\sqrt{1-4{{x}^{2}}}}\times \dfrac{d\left( 2x \right)}{dx}$ ---(6).
We know that $\dfrac{d\left( ax \right)}{dx}=a$. Let us use this result in equation (6).
$\Rightarrow \dfrac{dy}{dx}=\dfrac{2}{\sqrt{1-4{{x}^{2}}}}$.
$\therefore $ We have found the derivative of the function $\arcsin \left( 2x \right)$ as $\dfrac{2}{\sqrt{1-4{{x}^{2}}}}$.
Note: Whenever we get this type of problem, we try to make use of chain rule to get a solution to the given problem. We should not forget to different $2x$ after performing equation (5) which is the common mistake done by students. We can also solve this problem by making use of the fact that \[\dfrac{d}{dx}\left( \arcsin \left( f\left( x \right) \right) \right)=\dfrac{\dfrac{d\left( f\left( x \right) \right)}{dx}}{\sqrt{1-{{\left( f\left( x \right) \right)}^{2}}}}\] to get the required answer. Similarly, we can expect problems to find the derivative of the function $y=\arctan \left( \log \left( 5x \right) \right)$.
Complete step-by-step answer:
According to the problem, we are asked to find the derivative of the function $\arcsin \left( 2x \right)$.
Let us assume $y=\arcsin \left( 2x \right)$ ---(1).
Let us assume \[2x=z\]. Let us substitute this in equation (1).
$\Rightarrow y=\arcsin \left( z \right)$ ---(2).
Let us differentiate both sides of the equation (2) with respect to x.
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d\left( \arcsin \left( z \right) \right)}{dx}$ ---(3).
From chain rule of differentiation, we know that $\dfrac{d\left( g\left( f \right) \right)}{dx}=\dfrac{d\left( g \right)}{df}\times \dfrac{df}{dx}$. Let us substitute this result in equation (3).
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d\left( \arcsin \left( z \right) \right)}{dz}\times \dfrac{dz}{dx}$ ---(4).
We know that $\dfrac{d\left( \arcsin \left( x \right) \right)}{dx}=\dfrac{1}{\sqrt{1-{{x}^{2}}}}$. Let us use this result in equation (4).
$\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\sqrt{1-{{z}^{2}}}}\times \dfrac{dz}{dx}$ ---(5).
Now, let us substitute $z=2x$ in equation (5).
$\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\sqrt{1-4{{x}^{2}}}}\times \dfrac{d\left( 2x \right)}{dx}$ ---(6).
We know that $\dfrac{d\left( ax \right)}{dx}=a$. Let us use this result in equation (6).
$\Rightarrow \dfrac{dy}{dx}=\dfrac{2}{\sqrt{1-4{{x}^{2}}}}$.
$\therefore $ We have found the derivative of the function $\arcsin \left( 2x \right)$ as $\dfrac{2}{\sqrt{1-4{{x}^{2}}}}$.
Note: Whenever we get this type of problem, we try to make use of chain rule to get a solution to the given problem. We should not forget to different $2x$ after performing equation (5) which is the common mistake done by students. We can also solve this problem by making use of the fact that \[\dfrac{d}{dx}\left( \arcsin \left( f\left( x \right) \right) \right)=\dfrac{\dfrac{d\left( f\left( x \right) \right)}{dx}}{\sqrt{1-{{\left( f\left( x \right) \right)}^{2}}}}\] to get the required answer. Similarly, we can expect problems to find the derivative of the function $y=\arctan \left( \log \left( 5x \right) \right)$.
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