Question

If $y = {\sin ^{ - 1}}\left[ {\dfrac{{\left( {\sin \alpha \sin x} \right)}}{{\left( {1 - \cos \alpha \sin x} \right)}}} \right]$ then\[y'(0)\] is equal to
1.\[1\]
2. \[\tan \alpha \]
3. \[\left( {\dfrac{1}{2}} \right)\tan \alpha \]
4. \[\sin \alpha \]

Answer 1

Hint: In order to find the value of \[y'(0)\] find the first order derivative by differentiating the given function with respect to x. The function can be solved by using the Chain rule, which is as solving the inner function then solving the outer function, which is numerically written as:
$\dfrac{{dy}}{{dx}} = \dfrac{{du}}{{dc}} \times \dfrac{{dc}}{{dv}} \times \dfrac{{dv}}{{dx}}$, where u, c, v can be the functions inside the main function.
Formula used:
\[\dfrac{d}{{dx}}\left[ {\dfrac{{f(x)}}{{g(x)}}} \right] = \dfrac{{g(x)f'(x) - f(x)g'(x)}}{{{{\left[ {g(x)} \right]}^2}}}\]

Complete answer:
We are given with a function $y = {\sin ^{ - 1}}\left[ {\dfrac{{\left( {\sin \alpha \sin x} \right)}}{{\left( {1 - \cos \alpha \sin x} \right)}}} \right]$.
Since, the value inside the braces is in the form of division so, we can use the Quotient rule of derivatives when needed, which is:
\[\dfrac{d}{{dx}}\left[ {\dfrac{{f(x)}}{{g(x)}}} \right] = \dfrac{{g(x)f'(x) - f(x)g'(x)}}{{{{\left[ {g(x)} \right]}^2}}}\]
Starting with differentiation.
Considering $u = \dfrac{{\left( {\sin \alpha \sin x} \right)}}{{\left( {1 - \cos \alpha \sin x} \right)}}$
And, so, the value becomes:
$y = {\sin ^{ - 1}}u$
Differentiating u with respect to x using the quotient rule, we get:
Comparing:
\[
  f\left( x \right) = \left( {\sin \alpha \sin x} \right) \\
  g\left( x \right) = \left( {1 - \cos \alpha \sin x} \right) \\
 \]
And, substituting the values:
$\dfrac{{du}}{{dx}} = \dfrac{{\left( {1 - \cos \alpha \sin x} \right)\dfrac{{d\left( {\sin \alpha \sin x} \right)}}{{dx}} - \left( {\sin \alpha \sin x} \right)\dfrac{{d\left( {1 - \cos \alpha \sin x} \right)}}{{dx}}}}{{{{\left( {1 - \cos \alpha \sin x} \right)}^2}}}$
$ \Rightarrow \dfrac{{du}}{{dx}} = \dfrac{{\sin \alpha \left( {1 - \cos \alpha \sin x} \right)\dfrac{{d\left( {\sin x} \right)}}{{dx}} - \left( {\sin \alpha \sin x} \right)\left( {\dfrac{{d1}}{{dx}} - \cos \alpha \dfrac{{d\left( {\sin x} \right)}}{{dx}}} \right)}}{{{{\left( {1 - \cos \alpha \sin x} \right)}^2}}}$
Since, we know that $\dfrac{{d\left( {\sin x} \right)}}{{dx}} = \cos x$
so, we get:
$ \Rightarrow \dfrac{{du}}{{dx}} = \dfrac{{\sin \alpha \left( {1 - \cos \alpha \sin x} \right)\cos x - \left( {\sin \alpha \sin x} \right)\left( { - \cos x\cos \alpha } \right)}}{{{{\left( {1 - \cos \alpha \sin x} \right)}^2}}}$
$ \Rightarrow \dfrac{{du}}{{dx}} = \dfrac{{\sin \alpha \cos x\left( {1 - \cos \alpha \sin x} \right) + \left( {\sin \alpha \sin x\cos x\cos \alpha } \right)}}{{{{\left( {1 - \cos \alpha \sin x} \right)}^2}}}$
Taking $\sin \alpha \cos x$ common, we get:
$ \Rightarrow \dfrac{{du}}{{dx}} = \dfrac{{\sin \alpha \cos x\left( {1 - \cos \alpha \sin x + \sin x\cos \alpha } \right)}}{{{{\left( {1 - \cos \alpha \sin x} \right)}^2}}}$
$ \Rightarrow \dfrac{{du}}{{dx}} = \dfrac{{\sin \alpha \cos x}}{{{{\left( {1 - \cos \alpha \sin x} \right)}^2}}}$
 Now, differentiating both sides of the function y, with respect to \[x\] we get:
Since, we have $y = {\sin ^{ - 1}}u$, So, we can write:
$\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \times \dfrac{{du}}{{dx}}$
As, we know that $\dfrac{{d\left( {{{\sin }^{ - 1}}x} \right)}}{{dx}} = \dfrac{1}{{\sqrt {1 - {x^2}} }}$: So, we get:
\[ \Rightarrow y' = \dfrac{1}{{\sqrt {1 - {u^2}} }}\dfrac{{du}}{{dx}}\]
Substituting the value of u and $\dfrac{{du}}{{dx}}$ we get:
\[ \Rightarrow y' = \dfrac{1}{{\sqrt {1 - \dfrac{{{{\left( {\sin \alpha \sin x} \right)}^2}}}{{{{\left( {1 - \cos \alpha \sin x} \right)}^2}}}} }} \times \dfrac{{\sin \alpha \cos x}}{{{{\left( {1 - \cos \alpha \sin x} \right)}^2}}}\]
Solving it further, we get:
\[ \Rightarrow y' = \dfrac{1}{{\sqrt {\dfrac{{{{\left( {1 - \cos \alpha \sin x} \right)}^2} - {{\left( {\sin \alpha \sin x} \right)}^2}}}{{{{\left( {1 - \cos \alpha \sin x} \right)}^2}}}} }} \times \dfrac{{\sin \alpha \cos x}}{{{{\left( {1 - \cos \alpha \sin x} \right)}^2}}}\]
\[ \Rightarrow y' = \dfrac{{\left( {1 - \cos \alpha \sin x} \right)}}{{\sqrt {{{\left( {1 - \cos \alpha \sin x} \right)}^2} - {{\left( {\sin \alpha \sin x} \right)}^2}} }} \times \dfrac{{\sin \alpha \cos x}}{{{{\left( {1 - \cos \alpha \sin x} \right)}^2}}}\]
Cancelling \[\left( {1 - \cos \alpha \sin x} \right)\] from the numerator and denominator, we get:
\[ \Rightarrow y' = \dfrac{{\sin \alpha \cos x}}{{\left( {1 - \cos \alpha \sin x} \right)\sqrt {{{\left( {1 - \cos \alpha \sin x} \right)}^2} - {{\left( {\sin \alpha \sin x} \right)}^2}} }}\]
As, we needed to find the value of y=0:
So, now putting \[x = 0\] we get:
\[ \Rightarrow y' = \dfrac{{\sin \alpha \cos 0}}{{\left( {1 - \cos \alpha \sin 0} \right)\sqrt {{{\left( {1 - \cos \alpha \sin 0} \right)}^2} - {{\left( {\sin \alpha \sin 0} \right)}^2}} }}\]
Since, we know that \[\cos 0 = 1{\text{ and sin0 = 0}}\] so, we get:
\[ \Rightarrow y' = \dfrac{{\sin \alpha \times 1}}{{\left( {1 - 0} \right)\sqrt {{{\left( {1 - 0} \right)}^2} - {0^2}} }}\]
\[ \Rightarrow y' = \dfrac{{\sin \alpha \times 1}}{{1\sqrt {{1^2}} }}\]
\[ \Rightarrow y' = \sin \alpha \]
That gives the value \[y'(0) = \sin \alpha \].
Hence, Option (4) is the correct Answer.

Note:
If needed we can solve the obtained first order derivative more to its simplest form using the formula \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\], but since, we can see there would be no change in the answer after solving it further as we can substitute the value at this stage only.