Question

If ${{\left( \sin y \right)}^{\sin \left( \dfrac{\pi x}{2} \right)}}-\dfrac{\sqrt{3}}{2}{{\sec }^{-1}}\left( 2x \right)+{{2}^{x}}\tan \left( \log \left( x+2 \right) \right)=0$, then $\dfrac{dy}{dx}$ at x = -1 is
(a) $\dfrac{3}{\sqrt{{{\pi }^{2}}-3}}$
(b) $\dfrac{1}{\pi \sqrt{{{\pi }^{2}}-3}}$
(c) $\dfrac{3}{\pi \sqrt{{{\pi }^{2}}-3}}$
(d) $\dfrac{3\pi }{\sqrt{{{\pi }^{2}}-3}}$

Answer 1

Hint:Here take the parts of the given equations as A,B , C and differentiate them separately by using product and chain rule of differentiation.After differentiating find the value of $\dfrac{dy}{dx}$ at x = -1 and find the y value at x=-1 from the given equation and again substituting we get required value i.e $\dfrac{dy}{dx}$.

Complete step-by-step answer:
We are given that
${{\left( \sin y \right)}^{\sin \left( \dfrac{\pi x}{2} \right)}}-\dfrac{\sqrt{3}}{2}{{\sec }^{-1}}\left( 2x \right)+{{2}^{x}}\tan \left( \log \left( x+2 \right) \right)=0....\left( i \right)$
We have to find $\dfrac{dy}{dx}\text{ at }x=-1$.
Let $A={{\left( \sin y \right)}^{\sin \left( \dfrac{\pi x}{2} \right)}}.....\left( ii \right)$
By taking log on both sides of equation (i),
We get, $\log A=\log {{\left( \sin y \right)}^{\sin \left( \dfrac{\pi x}{2} \right)}}$
Since, we know that $\log {{m}^{n}}=n\log m$
Therefore, we get,
$\log A=\sin \left( \dfrac{\pi x}{2} \right).\log \left( \sin y \right)....\left( iii \right)$
Since, we know that $\dfrac{d}{dx}\left( \log x \right)=\dfrac{1}{x}$
$\dfrac{d}{dx}\left( \sin x \right)=\cos x$
By product rule, $\dfrac{d}{dx}\left( f\left( x \right).g\left( x \right) \right)=f\left( x \right).\dfrac{d\left( g\left( x \right) \right)}{dx}+g\left( x \right).\dfrac{d\left( f\left( x \right) \right)}{dx}....\left( iv \right)$
Also, by chain rule, we know that if $y=f\left( u \right)\text{ and }u=g\left( x \right)$
Then, $\dfrac{dy}{dx}={{f}^{'}}\left( u \right).\dfrac{du}{dx}....\left( v \right)$
By differentiating equation (iii) on both sides with respect to x, we get,
$\dfrac{1}{A}.\dfrac{dA}{dx}=\sin \left( \dfrac{\pi x}{2} \right).\dfrac{d}{dx}\left( \log \left( \sin y \right) \right)+\log \left( \sin y \right).\dfrac{d}{dx}\left( \sin \left( \dfrac{\pi x}{2} \right) \right)$
$\dfrac{1}{A}.\dfrac{dA}{dx}=\sin \left( \dfrac{\pi x}{2} \right).\left[ \dfrac{1}{\sin y}.\cos y \right].\dfrac{dy}{dx}+\log \left( \sin y \right).\cos \left( \dfrac{\pi x}{2} \right).\dfrac{\pi }{2}$
Since, $\dfrac{\cos x}{\sin x}=\cot x$
We get, $\dfrac{1}{A}.\dfrac{dA}{dx}=\sin \left( \dfrac{\pi x}{2} \right)\left( \cot y \right).\dfrac{dy}{dx}+\log \left( \sin y \right).\cos \left( \dfrac{\pi x}{2} \right).\dfrac{\pi }{2}$
By putting the value of A and cross multiplying, we get, $\dfrac{dA}{dx}={{\left( \sin y \right)}^{\sin \left( \dfrac{\pi x}{2} \right)}}\left[ \left( \left( \sin \dfrac{\pi x}{2} \right).\left( \cot y \right)\dfrac{dy}{dx} \right)+\log \left( \sin y \right).\cos \left( \dfrac{\pi x}{2} \right).\dfrac{\pi }{2} \right]$
Now, let $B=\dfrac{-\sqrt{3}}{2}{{\sec }^{-1}}\left( 2x \right)$
We know that $\dfrac{d}{dx}\left( {{\sec }^{-1}}x \right)=\dfrac{1}{x\sqrt{{{x}^{2}}-1}}$
Now, we will differentiate ‘B’ with respect to x by using chain rule from equation (v).
We get, $\dfrac{dB}{dx}=\dfrac{-\sqrt{3}}{2}\dfrac{1}{\left( 2x \right)\sqrt{{{\left( 2x \right)}^{2}}-1}}.2$
$\dfrac{dB}{dx}=\dfrac{-\sqrt{3}}{\left( 2x \right)\sqrt{4{{x}^{2}}-1}}$
Now, let $C={{2}^{x}}\tan \left( \log \left( x+2 \right) \right)$
We know that, $\dfrac{d}{dx}\left( {{a}^{x}} \right)={{a}^{x}}\log a$
$\dfrac{d}{dx}\left( \tan x \right)={{\sec }^{2}}x$
$\dfrac{d}{dx}\left( \log x \right)=\dfrac{1}{x}$
Now, we will differentiate ‘C’ by using product rule from equation (iv).
We get, $\dfrac{dC}{dx}=\tan \left( \log \left( x+2 \right) \right).\dfrac{d}{dx}\left( {{2}^{x}} \right)+{{2}^{x}}.\dfrac{d}{dx}\left( \tan \left( \log \left( x+2 \right) \right) \right)$
By using chain rule from equation (iv), we get, $\dfrac{dC}{dx}=\left[ \left( {{2}^{x}}\log 2 \right).\tan \left( \log \left( x+2 \right) \right) \right]+{{2}^{x}}.{{\sec }^{2}}\left( \log \left( x+2 \right) \right).\dfrac{1}{x+2}$
$\Rightarrow \dfrac{dC}{dx}=\left( {{2}^{x}}\log 2 \right).\tan \left( \log \left( x+2 \right) \right)+\dfrac{{{2}^{x}}{{\sec }^{2}}\left[ \log \left( x+2 \right) \right]}{\left( x+2 \right)}$
Now, we can write equation (i) as
$A+B+C=0$
By differentiating the equation, we get,
$\dfrac{dA}{dx}+\dfrac{dB}{dx}+\dfrac{dC}{dx}=0$
By putting the values of $\dfrac{dA}{dx},\dfrac{dB}{dx}\text{and}\dfrac{dC}{dx}$. We get, $\begin{aligned} & {{\left( \sin y \right)}^{\sin \left( \dfrac{\pi x}{2} \right)}}\left[ \sin \left( \dfrac{\pi x}{2} \right).\cot y.\dfrac{dy}{dx} \right]+\left[ \log \left( \sin y \right).\cos \left( \dfrac{\pi x}{2} \right).\dfrac{\pi }{2} \right]+\left[ \dfrac{-\sqrt{3}}{\left( 2x \right)\sqrt{4{{x}^{2}}-1}} \right] \\\ & +\left[ \left( {{2}^{x}}\log 2 \right).\tan \left( \log \left( x+2 \right) \right) \right]+\dfrac{{{2}^{x}}{{\sec }^{2}}\left[ \log \left( x+2 \right) \right]}{\left( x+2 \right)}=0 \\\ \end{aligned}$
Now to get $\dfrac{dy}{dx}\text{ at }x=-1$, we will put x = -1 in the above equation

& {{\left( \sin y \right)}^{\sin \left( \dfrac{\pi \left( -1 \right)}{2} \right)}}\left[ \sin \left( \dfrac{\pi \left( -1 \right)}{2} \right).\cot y.\dfrac{dy}{dx} \right]+\left[ \log \left( \sin y \right).\cos \left( \dfrac{-\pi }{2} \right).\dfrac{\pi }{2} \right] \\\ & +\left( \dfrac{-\sqrt{3}}{\left( 2\left( -1 \right) \right)\sqrt{4{{\left( -1 \right)}^{2}}-1}} \right)+\left[ \left( {{2}^{-1}}\log 2 \right).\tan \left( \log \left( 2-1 \right) \right) \right]+\dfrac{{{2}^{-1}}{{\sec }^{2}}\left[ \log \left( 2-1 \right) \right]}{\left( 2-1 \right)}=0 \\\ \end{aligned}$$ Since we know that $$\sin \left( \dfrac{-\pi }{2} \right)=-\sin \dfrac{\pi }{2}=-1\text{ and }\cos \left( \dfrac{-\pi }{2} \right)=0$$ We get, $${{\left( \sin y \right)}^{-1}}\left[ \left( -1 \right).\cot y.\dfrac{dy}{dx} \right]+0+\left( \dfrac{-\sqrt{3}}{\left( -2 \right)\sqrt{3}} \right)+\left[ \left( {{2}^{-1}}.\log 2 \right).\tan \left( \log \left( 1 \right) \right) \right]+\dfrac{{{2}^{-1}}{{\sec }^{2}}\left[ \log \left( 1 \right) \right]}{\left( 1 \right)}=0$$ Since, we know that log 1 = 0, therefore, we get $$\tan \left( \log \left( 1 \right) \right)=\tan {{0}^{0}}=0\text{ and }{{\sec }^{2}}\left( \log \left( 1 \right) \right)={{\sec }^{2}}0=1$$ Therefore, we get, $${{\left( \sin y \right)}^{-1}}\left[ \left( -\cot y \right)\dfrac{dy}{dx} \right]+\dfrac{1}{2}+0+\dfrac{{{2}^{-1}}}{1}=0$$ $$={{\left( \sin y \right)}^{-1}}\left( -\cot y \right).\dfrac{dy}{dx}+\dfrac{1}{2}+\dfrac{1}{2}=0$$ $$=-{{\left( \sin y \right)}^{-1}}\left( \cot y \right)\dfrac{dy}{dx}+1=0....\left( vi \right)$$ Now to find $$\dfrac{dy}{dx}\text{ at }x=-1$$, we have to find $${{\left( \sin y \right)}^{-1}}$$ and $$\cot y$$ at x = -1. For that we put x = -1 in equation (i), we get, $${{\left( \sin y \right)}^{\sin \left( \dfrac{-\pi }{2} \right)}}\dfrac{-\sqrt{3}}{2}{{\sec }^{-1}}\left( -2 \right)+{{2}^{-1}}\tan \left( \log \left( 2-1 \right) \right)=0$$ Since, we know that $${{\sec }^{-1}}\left( -2 \right)=\dfrac{2\pi }{3}$$ We get, $${{\left( \sin y \right)}^{-1}}\dfrac{-\sqrt{3}}{2}\left( \dfrac{2\pi }{3} \right)+0=0$$ $${{\left( \sin y \right)}^{-1}}=\dfrac{\pi }{\sqrt{3}}$$ $$\dfrac{1}{\sin y}=\dfrac{\pi }{\sqrt{3}}$$ $$\dfrac{\sqrt{3}}{\pi }=\sin y$$ Since, we know that $$\cos y=\sqrt{1-{{\sin }^{2}}y}$$ We get, $$\cos y=\sqrt{1-{{\left( \dfrac{\sqrt{3}}{\pi } \right)}^{2}}}$$ $$\cos y=\sqrt{1-\dfrac{3}{{{\pi }^{2}}}}$$ Therefore, $$\cot y=\dfrac{\cos y}{\sin y}=\dfrac{\sqrt{1-\dfrac{3}{{{\pi }^{2}}}}}{\dfrac{\sqrt{3}}{\pi }}$$ We get, $$\cot y=\dfrac{\sqrt{{{\pi }^{2}}-3}}{\sqrt{3}}$$ Now, we will put the values of $${{\left( \sin y \right)}^{-1}}$$ and $${{\cot }^{-1}}y$$ in equation (vi), we get $$-\left( \dfrac{\pi }{\sqrt{3}} \right)\left( \dfrac{\sqrt{{{\pi }^{2}}-3}}{\sqrt{3}} \right)\dfrac{dy}{dx}+1=0$$ $$\dfrac{dy}{dx}=\dfrac{-1}{\dfrac{-\left( \pi \right)\left( \sqrt{{{\pi }^{2}}-3} \right)}{3}}$$ So we get, $$\dfrac{dy}{dx}=\dfrac{3}{\left( \pi \right)\left( \sqrt{{{\pi }^{2}}-3} \right)}$$ Hence, the correct option is (c). Note: In such a type of question, students should always try to break the question in parts and then differentiate the equation to avoid confusion and mistakes. Also take special care of signs in equations.Remember the formulas of differentiation of trigonometric ,inverse trigonometric and logarithmic functions and also formula of differentiation of product rule and quotient rule to solve these types of questions.