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# Differentiate with respect to $x:\sin \left( m{{\sin }^{-1}}x \right)$(a) $\cos \left( m{{\cos }^{-1}}x \right)$(b) $\sin \left( m{{\sin }^{-1}}x \right)$(c) ${{m}^{2}}\sin x$(d) none of these

Last updated date: 14th Jul 2024
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Hint: To solve this question, we can use chain rule since we have to differentiate a composite function of the form $f\left( g\left( x \right) \right)$.

In this question, we have to differentiate $\sin \left( m{{\sin }^{-1}}x \right)$ with respect to
$x$. Before proceeding with the question, we must know the chain rule. If we have to differentiate a function which is the form of $f\left( g\left( x \right) \right)$, we will use chain rule. We can differentiate a function which is the form of $f\left( g\left( x \right) \right)$ using chain rule as shown below,
$\dfrac{d\left( f\left( g\left( x \right) \right) \right)}{dx}=\dfrac{d\left( f\left( g\left( x \right) \right) \right)}{d\left( g\left( x \right) \right)}\times \dfrac{d\left( g\left( x \right) \right)}{dx}.............\left( 1 \right)$
In the question, since we are given a function $f\left( g\left( x \right) \right)=\sin \left( m{{\sin }^{- 1}}x \right)$. So, we can find out $g\left( x \right)=m{{\sin }^{-1}}x$. Substituting $f\left( g\left( x \right) \right)=\sin \left( m{{\sin }^{-1}}x \right)$ and $g\left( x \right)=m{{\sin }^{-1}}x$ in equation
$\left( 1 \right)$, we get,
$\dfrac{d\left( \sin \left( m{{\sin }^{-1}}x \right) \right)}{dx}=\dfrac{d\left( \sin \left( m{{\sin }^{-1}}x \right) \right)}{d\left( m{{\sin }^{-1}}x \right)}\times \dfrac{d\left( m{{\sin }^{-1}}x \right)}{dx}.............\left( 2 \right)$
Since we are differentiating $\sin \left( m{{\sin }^{-1}}x \right)$ with respect to $m{{\sin }^{-1}}x$, we
get,
$\dfrac{d\left( \sin \left( m{{\sin }^{-1}}x \right) \right)}{d\left( m{{\sin }^{-1}}x \right)}=\cos \left( m{{\sin }^{-1}}x \right)$, $\because \dfrac{d\sin x}{dx}=\cos x$
Also, we have a formula which gives us the derivative of ${{\sin }^{-1}}x$ with respect to $x$,
$\dfrac{d{{\sin }^{-1}}x}{dx}=\dfrac{1}{\sqrt{1-{{x}^{2}}}}..........\left( 4 \right)$
Since $m$is a constant, we can take $m$out of the derivative in $\dfrac{d\left( m{{\sin }^{-1}}x \right)}{dx}$ and hence, we can write $\dfrac{d\left( m{{\sin }^{-1}}x \right)}{dx}$ as,
$\dfrac{d\left( m{{\sin }^{-1}}x \right)}{dx}=m\dfrac{d\left( {{\sin }^{-1}}x \right)}{dx}$

Substituting $\dfrac{d{{\sin }^{-1}}x}{dx}=\dfrac{1}{\sqrt{1-{{x}^{2}}}}$ from equation $\left( 4 \right)$ in the above equation to obtain the derivative of $m{{\sin }^{-1}}x$ with respect to $x$, we
get,
$\dfrac{d\left( m{{\sin }^{-1}}x \right)}{dx}=\dfrac{m}{\sqrt{1-{{x}^{2}}}}...........\left( 5 \right)$
Substituting $\dfrac{d\left( \sin \left( m{{\sin }^{-1}}x \right) \right)}{d\left( m{{\sin }^{-1}}x \right)}=\cos \left( m{{\sin }^{-1}}x \right)$ from equation $\left( 3 \right)$ and $\dfrac{d\left( m{{\sin }^{-1}}x \right)}{dx}=\dfrac{m}{\sqrt{1-{{x}^{2}}}}$ from equation $\left( 5 \right)$ in
equation $\left( 2 \right)$, we get,
\begin{align} & \dfrac{d\left( \sin \left( m{{\sin }^{-1}}x \right) \right)}{dx}=\cos \left( m{{\sin }^{-1}}x \right)\times \dfrac{m}{\sqrt{1-{{x}^{2}}}} \\ & \Rightarrow \dfrac{d\left( \sin \left( m{{\sin }^{-1}}x \right) \right)}{dx}=\dfrac{m\cos \left( m{{\sin }^{-1}}x \right)}{\sqrt{1-{{x}^{2}}}} \\ \end{align}
None of the options are matching with the answer that is $\dfrac{m\cos \left( m{{\sin }^{-1}}x \right)}{\sqrt{1-{{x}^{2}}}}$.

Therefore the correct answer is option (d).

Note: There is a possibility of committing a mistake while writing the derivative of $\sin x$ or $\cos x$. There is always a confusion in writing the negative sign. One may write the derivative of $\sin x$ as $-\cos x$ instead of $\cos x$. Also, one may write the derivative of $\cos x$ as $\sin x$ instead of $- \sin x$. So, in order to avoid such mistakes, one must remember them thoroughly.