Question

solve this question otherwise don't try​

Answer 1

Hey Stark you've got a fan here ^_^

As for your question we have,

$\frac{d}{dx} [\frac{x}{2}\sqrt{a^{2} -x^{2} } + \frac{a^{2} }{2}sin^{-1}\frac{x}{a} ]$

I'll be doing as always the LHS part first,

and

Let's do the simpler parts first,

i.e Splitting,

Which gives,

$\frac{d}{dx} [\frac{x}{2}\sqrt{a^{2} -x^{2} } ]+ \frac{d}{dx}[ \frac{a^{2} }{2}sin^{-1}\frac{x}{a} ]$

Now just for the sake of not making this equation too messy,

Let me just solve the two terms alone first and then then bring them back together in the end,

The first term is,

$\frac{d}{dx} [\frac{x}{2}\sqrt{a^{2} -x^{2} } ]$

Using relation,

$\frac{dxy}{dx} = y\frac{dx}{dx} +x\frac{dy}{dx}$

We get,

$(\sqrt{a^{2} -x^{2} } )\frac{d}{dx} (\frac{x}{2}) +( \frac{x}{2}) \frac{d}{dx}( \sqrt{a^{2} -x^{2} })\\\\=(\sqrt{a^{2} -x^{2} })( \frac{1}{2} )+ \frac{x}{2} (-\frac{x}\sqrt{a^{2} -x^{2} } } )\\\\= \frac{a^{2}-x^{2} - x^{2} }{2\sqrt{a^{2} -x^{2} } } \\\\=\frac{a^{2}-2x^{2} }{2\sqrt{a^{2} -x^{2} } }$

Now time for the second part i.e,

$\frac{d}{dx}[ \frac{a^{2} }{2}sin^{-1}\frac{x}{a} ]\\\\\\=(\frac{a^{2} }{2}) \frac{d}{dx} (sin^{-1}\frac{x}{a} ) +(sin^{-1} \frac{x}{a}) \frac{d}{dx} (\frac{a^{2} }{2} )\\\\=\frac{a^{2} }{2} (\frac{1}{\sqrt{1-\frac{x^{2} }{a^{2} } } }) (\frac{1}{a}) \\= \frac{a^{2} }{2\sqrt{a^{2}-x^{2} } }$

since,

$sin^{-1} \frac{x}{a} = (\frac{1}{1-\frac{x^{2} }{a^{2} } } )(\frac{1}{a} )$

Now that we have our two parts let's combine them like a brave mathematician would,

$=\frac{a^{2}-2x^{2} }{2\sqrt{a^{2} -x^{2} } } + \frac{a^{2} }{2\sqrt{a^{2}-x^{2} } }\\=\frac{2a^{2}-2x^{2} }{2\sqrt{a^{2} -x^{2} } } \\=\frac{a^{2}-x^{2} }{\sqrt{a^{2} -x^{2} } }$

Rationalizing the terms gives,

$\frac{(a^{2} -x^{2} )(\sqrt{a^{2} -x^{2} } )}{(\sqrt{a^{2} -x^{2} } )(\sqrt{a^{2} -x^{2} } )} \\=\frac{(a^{2} -x^{2}) \sqrt{a^{2} -x^{2} } }{(a^{2} -x^{2} )} \\=\sqrt{a^{2} -x^{2}} =RHS$

Hence proved.

huh took a lot of effort typing this thing.

And BTW miss you 3000.

Answer 2

Answer:

ya hai answer........