
How do you find the derivative of $\arcsin \left( 2{{x}^{2}} \right)$ ?
Answer
556.8k+ views
Hint: In this question, we have to find the derivative of a inverse trigonometric function. So, we will use the formulas for the derivative of a trigonometric function, to get the result. We first let $2{{x}^{2}}=u$ and thus, find the derivative of the new trigonometric function with respect to x. After that, we will again find the derivative of $2{{x}^{2}}=u$with respect to x and will put the value of $\dfrac{du}{dx}$ in the equation, to get the required result to the problem.
Complete step-by-step solution:
According to the question, we have to find the derivative of a trigonometric function.
So, we will apply the trigonometric derivative formulas to get the result.
The trigonometric function given to us is $\arcsin \left( 2{{x}^{2}} \right)$ --------- (1)
So, we will first let $2{{x}^{2}}=u$ in the equation (1), we get
$y=\arcsin (u)$
So, now we will find the derivative of the above equation with respect to x, which is derivative of sin inverse is ${{\left( {{\sin }^{-1}}x \right)}^{\prime }}=\dfrac{1}{\sqrt{1-{{x}^{2}}}}$ we get
$\dfrac{dy}{dx}=\dfrac{1}{\sqrt{1-{{u}^{2}}}}.\dfrac{du}{dx}$ ---------- (2)
Now, we will find the derivative of the equation $2{{x}^{2}}=u$ with respect to x, which is
$u=2{{x}^{2}}$
$\left( \dfrac{du}{dx} \right)=4x$ ------- (3)
So, we will put the value of u and its derivative, that is the value equation (4) in equation (3), we get
$\dfrac{dy}{dx}=\dfrac{1}{\sqrt{1-{{\left( 2{{x}^{2}} \right)}^{2}}}}.(4x)$
On further simplification, we get
$\dfrac{dy}{dx}=\dfrac{4x}{\sqrt{1-4{{x}^{4}}}}$ which is our required solution to the problem.
Therefore, for the trigonometric function $\arcsin \left( 2{{x}^{2}} \right)$ , its derivative is equal to $\dfrac{4x}{\sqrt{1-4{{x}^{4}}}}$.
Note: While calculating the answer to the problem, do mention the derivations properly to avoid confusion and mathematical errors. One of the alternative methods to solve this problem is instead of letting $u=2{{x}^{2}}$, solve the trigonometric function as the whole, that is first find the derivative of ${{\sin }^{-1}}$ and then find the derivative of $2{{x}^{2}}$ and make further simplification, to get the required result for the problem.
Complete step-by-step solution:
According to the question, we have to find the derivative of a trigonometric function.
So, we will apply the trigonometric derivative formulas to get the result.
The trigonometric function given to us is $\arcsin \left( 2{{x}^{2}} \right)$ --------- (1)
So, we will first let $2{{x}^{2}}=u$ in the equation (1), we get
$y=\arcsin (u)$
So, now we will find the derivative of the above equation with respect to x, which is derivative of sin inverse is ${{\left( {{\sin }^{-1}}x \right)}^{\prime }}=\dfrac{1}{\sqrt{1-{{x}^{2}}}}$ we get
$\dfrac{dy}{dx}=\dfrac{1}{\sqrt{1-{{u}^{2}}}}.\dfrac{du}{dx}$ ---------- (2)
Now, we will find the derivative of the equation $2{{x}^{2}}=u$ with respect to x, which is
$u=2{{x}^{2}}$
$\left( \dfrac{du}{dx} \right)=4x$ ------- (3)
So, we will put the value of u and its derivative, that is the value equation (4) in equation (3), we get
$\dfrac{dy}{dx}=\dfrac{1}{\sqrt{1-{{\left( 2{{x}^{2}} \right)}^{2}}}}.(4x)$
On further simplification, we get
$\dfrac{dy}{dx}=\dfrac{4x}{\sqrt{1-4{{x}^{4}}}}$ which is our required solution to the problem.
Therefore, for the trigonometric function $\arcsin \left( 2{{x}^{2}} \right)$ , its derivative is equal to $\dfrac{4x}{\sqrt{1-4{{x}^{4}}}}$.
Note: While calculating the answer to the problem, do mention the derivations properly to avoid confusion and mathematical errors. One of the alternative methods to solve this problem is instead of letting $u=2{{x}^{2}}$, solve the trigonometric function as the whole, that is first find the derivative of ${{\sin }^{-1}}$ and then find the derivative of $2{{x}^{2}}$ and make further simplification, to get the required result for the problem.
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