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My attempt:

$\eqalign{ & x = 4\sin \left( {2y + 6} \right) \cr & {{dx} \over {dy}} = \left( 2 \right)\left( 4 \right)\cos \left( {2y + 6} \right) \cr & {{dx} \over {dy}} = 8\cos \left( {2y + 6} \right) \cr & {{dy} \over {dx}} = {1 \over {8\cos \left( {2y + 6} \right)}} \cr} $

$\eqalign{ & x = 4\sin \left( {2y + 6} \right) \cr & {{dx} \over {dy}} = \left( 2 \right)\left( 4 \right)\cos \left( {2y + 6} \right) \cr & {{dx} \over {dy}} = 8\cos \left( {2y + 6} \right) \cr & {{dy} \over {dx}} = {1 \over {8\cos \left( {2y + 6} \right)}} \cr & {\cos ^2}\left( {2y + 6} \right) + {\sin ^2}\left( {2y + 6} \right) = 1 \cr & {\cos ^2}\left( {2y + 6} \right) + {\left( {{x \over 4}} \right)^2} = 1 \cr & {\cos ^2}\left( {2y + 6} \right) = 1 - {{{x^2}} \over {16}} \cr & \cos \left( {2y + 6} \right) = \sqrt {{{16 - {x^2}} \over {16}}} \cr & \cos \left( {2y + 6} \right) = {{\sqrt {16 - {x^2}} } \over 4} \cr & {{dy} \over {dx}} = {1 \over {2\sqrt {16 - {x^2}} }} = 1 \cr} $

Okay I've got it right, but the official answer confuses me, it says:

${{dy} \over {dx}} = {1 \over {8cos\left( {\arcsin \left( {{x \over 4}} \right)} \right)}} = \left( {\left( \pm \right){1 \over {2\sqrt {\left( {16 - {x^2}} \right)} }}} \right)$

This is the part i'm struggling to get my head around, although I arrive at the same answer.

Okay

$\arcsin {x \over 4} = 2y + 6$

but how does the answer then go :

$\left( {\left( \pm \right){1 \over {2\sqrt {\left( {16 - {x^2}} \right)} }}} \right)$

is there a shortcut or trick I overlooked?

I think I need some sleep, thanks...

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3 Answers 3

up vote 3 down vote accepted

You got:

$\dfrac{dy}{dx}=\dfrac{1}{8\cos(2y+6)}$

And we have

$x=4\sin(2y+6)\implies y=\dfrac{1}{2}\left(\sin^{-1}\left(\dfrac{x}{4}\right)-6\right)$

Plug that in the top equation:

$\dfrac{dy}{dx}=\dfrac{1}{8\cos\left(\sin^{-1}\left(\dfrac{x}{4}\right)\right)}$

We know $\cos^2x+\sin^2x=1\implies \cos x=\pm\sqrt{1-\sin^2x}$

and therefore we say $\cos(\sin^{-1}x)=\pm\sqrt{1-x^2}$, we have

$\dfrac{dy}{dx}=\pm\dfrac{1}{8\sqrt{1-\dfrac{x^2}{16}}}\implies \dfrac{dy}{dx}=\pm\dfrac{1}{2\sqrt{16-x^2}}$

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You have $x=4\sin(2y+6)$, simply solve for $y$ to get $y=\dfrac{1}{2}\left(\sin^{-1}\left(\dfrac{x}{4}\right)-6\right)$. That's what I did. –  Maazul Jun 10 '13 at 0:34

To find $\frac{dy}{dx}$ in terms of $x$ for $x=4 \sin(2y+6)$, I wouldn't bother with implicit differentiation. Instead:

First, solve for $y$,

$$\frac{x}{4} = \sin(2y+6)$$

$$\arcsin(\frac{x}{4}) = 2y + 6$$

$$y = \frac{\arcsin(\frac{x}{4})-6}{2} = \frac{1}{2} \arcsin{\frac{x}{4}} - 3.$$

Now differentiate with respect to $x$,

$$\frac{dy}{dx} = \frac{1}{2} \frac{1}{\sqrt{16-x^2}} = \frac{1}{2 \sqrt{16-x^2}}.$$

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I havent learned how to differentiate inverse functions yet.. –  Assad Jun 10 '13 at 0:25
    
Then you may have to try using implicit differentiation, such as what @Maazul suggested. –  Gamma Function Jun 10 '13 at 0:27
2  
@Assad For future reference, here is the derivative of arcsin(x): $$y=\arcsin(x)$$ $$\sin(y)=x \ \ \ \ \ $$ (Note for the last line of the proof that $\sin(y)=x \implies \sin^2(y)=x^2$) $$cos(y) \frac{dy}{dx}=1$$ $$\frac{dy}{dx} = \frac{1}{\cos(y)}$$ Now, note that $$\sin^2(y)+ \cos^2(y) = 1 \implies \cos^2(y)=1-\sin^2(y) \implies \cos(y)=\sqrt{1-\sin^2(y)}$$ So, $$\frac{dy}{dx} = \frac{1}{\cos(y)} = \frac{1}{\sqrt{1-\sin^2(y)}} = \frac{1}{\sqrt{1-x^2}}.$$ –  Gamma Function Jun 10 '13 at 1:20

You are on the right track. Notice that $$\cos^2(2y+6) = 1-\frac{x^2}{16}$$ implies $$\cos(2y+6) = \pm\sqrt{1-\frac{x^2}{16}}.$$

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