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Mathematics 8 Online
OpenStudy (anonymous):

differentiate w.r.t x

Parth (parthkohli):

Please cite your sources @Rohangrr as lgba said

Parth (parthkohli):

To differentiate an expression: \(\Large x^n\), the formula is: \(\Large nx^{n - 1}\)

OpenStudy (anonymous):

(tan^-1){[(sqrt)(1+((a^2)(x^2)))]-1}/ax differentiate w.r.t x

OpenStudy (mimi_x3):

is it: \[\tan^{-1}(\frac{\sqrt{x}(1+x^{2}x^{2})-1}{ax}) \]?

OpenStudy (ash2326):

\[\tan^{-1}(\frac{\sqrt{1+a^2x^2}-1}{ax})\]???

OpenStudy (anonymous):

yea thats it

OpenStudy (ash2326):

\[\frac{d}{dx}(\tan^{-1} x)=\frac{1}{1+x^2}\] Using that here \[\frac{d}{dx} \tan^{-1} (\frac{\sqrt{1+a^2x^2}-1}{ax})=\frac{1}{1+(\frac{\sqrt{1+a^2x^2}-1}{ax})^2}\times \frac{d}{dx} (\frac{\sqrt{1+a^2x^2}-1}{ax})\] \[=\frac{1}{1+(\frac{\sqrt{1+a^2x^2}-1}{ax})^2}\times (\frac{ax \times \frac{d}{dx}(\sqrt{1+a^2x^2}-1)-(\sqrt{1+a^2x^2}-1)\frac{d}{dx}(ax)}{a^2x^2})\] Do you understand till here? after this only simplification is needed

OpenStudy (ash2326):

\[\small =\frac{1}{1+(\frac{\sqrt{1+a^2x^2}-1}{ax})^2}\times (\frac{ax \times \frac{d}{dx}(\sqrt{1+a^2x^2}-1)-(\sqrt{1+a^2x^2}-1)\frac{d}{dx}(ax)}{a^2x^2})\]

OpenStudy (anonymous):

i tried it be substituting x=tanx/a but after 3 steps dint no wat to do could you try substituting and then differentiationg it

OpenStudy (ash2326):

Ok, I'll do that

OpenStudy (anonymous):

yea thanks

OpenStudy (ash2326):

Oh no, the page restarted. Wait I'll type again

OpenStudy (anonymous):

okay

OpenStudy (ash2326):

Let's substitute \[x=\frac{\tan u}{a}\] \[y=\tan^{-1} (\frac{\sqrt{1+a^2x^2}-1}{ax})\] After substitution, we get \[y=\tan^{-1} (\frac{\sqrt {\sec^2u}-1}{\tan u})\]

OpenStudy (ash2326):

Do you get this part?

OpenStudy (anonymous):

yea

OpenStudy (ash2326):

We know \[1-\cos u= 2\sin^2 \frac u 2\] and \[\sin u= 2\sin \frac u2 \cos \frac u2\] Let's substitute these, we get \[y=\tan^{-1} (\frac{2 \sin^2 \frac u 2}{2\sin \frac u2 \cos \frac u2})\] So we get \[y=\tan ^{-1} ({\tan \frac u2})\]

OpenStudy (ash2326):

So we get \[y=\frac u 2 \] or \[\frac{dy}{du}=\frac 12\] We had substituted \[x=\frac {\tan u}{a}\] or \[u= \tan ^{-1} ax\] so \[\frac{du}{dx}= a \times \frac{1}{1+a^2x^2}\] I think you can do now, can't you?

OpenStudy (anonymous):

yea thanks

OpenStudy (ash2326):

Phew !! The page restarted so many times:)

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