Question

If $y=\sin (2{{\sin }^{-1}}x)$, then $\dfrac{dy}{dx}=$
A.$\dfrac{(2-4{{x}^{2}})}{\sqrt{(1-{{x}^{2}})}}$
B.$\dfrac{(2+4{{x}^{2}})}{\sqrt{(1-{{x}^{2}})}}$
C.$\dfrac{(2-4{{x}^{2}})}{\sqrt{(1+{{x}^{2}})}}$
D.None of these

Answer 1

Hint : To find the derivative we can use the slope formula, that is, $slope=\dfrac{dy}{dx}$.
$dy$is the changes in $Y$and $dx$ is the changes in $X$

To find the derivative of two variables in multiplication, this formula is used
$\dfrac{d}{dx}(uv)=u\dfrac{d}{dx}(v)+v\dfrac{d}{dx}(u)$
Mechanically, $\dfrac{d}{dx}(f(x))$ measures the rate of change of $f(x)$ with respect to $x$ .
Differentiation of any constant is zero. Differentiation of constant and a function is equal to constant times the differentiation of the function. Geometrically, graph of a constant function is a straight line parallel to the $x-axis$. Consequently slope of the tangent is zero.

Complete step-by-step answer :
Given that $y=\sin (2{{\sin }^{-1}}x)$
Let us assume that $x=\sin \theta$
Differentiating both sides with respect to $\theta$ we get
$\dfrac{dx}{d\theta }=\cos \theta$
Substituting these values in the equation we get
$y=\sin (2{{\sin }^{-1}}\sin \theta )$
Solving the above equation we get
$y=\sin 2\theta$
Differentiating both sides with respect to $\theta$ we get
$\dfrac{dy}{d\theta }=2\cos 2\theta$
We know that $\dfrac{dy}{dx}=\dfrac{dy}{d\theta }\times \dfrac{d\theta }{dx}$
Substituting the values in the equation we get
$\dfrac{dy}{dx}=\dfrac{2\cos 2\theta }{\cos \theta }$
Further solving by applying trigonometric identities we get
$\dfrac{dy}{dx}=\dfrac{2(1-2{{\sin }^{2}}\theta )}{\sqrt{(1-{{\sin }^{2}}\theta )}}$
Substituting $x=\sin \theta$ in the above equation we get
$\dfrac{dy}{dx}=\dfrac{2(1-2{{x}^{2}})}{\sqrt{1-{{x}^{2}}}}$
Further solving we get
$\dfrac{dy}{dx}=\dfrac{(2-4{{x}^{2}})}{\sqrt{(1-{{x}^{2}})}}$
Therefore, option $A$ is the correct answer.
So, the correct answer is “Option A”.

Note : $\dfrac{d}{dx}(f(x))=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(x+h)-f(x)}{h}$ is the formula for finding the derivative from the first principles. The slope is the rate of change of $y$ with respect to $x$ that means if $x$ is increased by an additional unit the change in $y$ is given by $\dfrac{dy}{dx}$ . Let us understand with an example, the rate of change of displacement of an object is defined as the velocity $Km/hr~$ that means when time is increased by one hour the displacement changes by $Km$. For solving derivative problems different techniques of differentiation must be known.