What is the second derivative of inverse tangent?

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The reply to this inquiry is ##1/(1+x^2)##. However, it's ameliorate to comprise why it's that. So near I accomplish teach why.

So we're involved to experience what the derivative of ##arctanx## is. Let's put it into a part.

##y=arctanx##

The proximate step is to resemble this heterogeneous. ##tany=x##

We can perchance implicitly verify this. We understand that ##d/dx(tan(u(x)))=sec^2(u(x))*u'(x)##

So near we implicitly verify, ##sec^2(y)*y'=1## We reresolve for ##y'##, and we get: ##y'=1/sec^2(y)##

But, is this what we are contemplateing for? Not indeed. It's nonentity obstruct to ##1/(1+x^2)##. We want past resources; fortunately, our trigonometric kindred frequently conclude beneficial during derivatives and integrations.

So let's use the one from the pythagorean theorem and appraise it this way: ##cos^2(y)+sin^2(y)=1## ##cos^2(y)/cos^2(y)+sin^2(y)/cos^2(y)=1/cos^2(y)## We use this so that we end up delay the secant part. => ##1+tan^2(y)=sec^2(y)##

So, let's supply this into our separation. => ##y'=1/sec^2(y)=1/(1+tan^2(y))##

Although obstructr, it tranquil does not contemplate love our separation, but do you recollect when we wrote the part heterogeneous? Isn't ##tany=x##?

So we would fair possess to supply it, and we get: ##d/dxarctan(x)=1/(1+x^2)##.

I expectation that this was beneficial, and now I canvass you to experience ##d/dxsec^-1(x)##. (inverse part of ##secx## and not ##1/secx##)

The best way to yield in Calculus is to indeed comprise and not memorize.

If you possess any inquiry, move clear to ask :).

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