differentiate w.r.t x
Please cite your sources @Rohangrr as lgba said
To differentiate an expression: \(\Large x^n\), the formula is: \(\Large nx^{n - 1}\)
(tan^-1){[(sqrt)(1+((a^2)(x^2)))]-1}/ax differentiate w.r.t x
is it: \[\tan^{-1}(\frac{\sqrt{x}(1+x^{2}x^{2})-1}{ax}) \]?
\[\tan^{-1}(\frac{\sqrt{1+a^2x^2}-1}{ax})\]???
yea thats it
\[\frac{d}{dx}(\tan^{-1} x)=\frac{1}{1+x^2}\] 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
\[\small =\frac{1}{1+(\frac{\sqrt{1+a^2x^2}-1}{ax})^2}\times (\frac{ax \times \frac{d}{dx}(\sqrt{1+a^2x^2}-1)-(\sqrt{1+a^2x^2}-1)\frac{d}{dx}(ax)}{a^2x^2})\]
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
Ok, I'll do that
yea thanks
Oh no, the page restarted. Wait I'll type again
okay
Let's substitute \[x=\frac{\tan u}{a}\] \[y=\tan^{-1} (\frac{\sqrt{1+a^2x^2}-1}{ax})\] After substitution, we get \[y=\tan^{-1} (\frac{\sqrt {\sec^2u}-1}{\tan u})\]
Do you get this part?
yea
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 \[y=\tan ^{-1} ({\tan \frac u2})\]
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?
yea thanks
Phew !! The page restarted so many times:)
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