Find derivative of tan-i (sin2x).

1.	Find derivative of tan-i (sin2x).
Answer» The derivative oftan−1xis11+x2(for "why", see note below) So, applying the chain rule, we get: ddx(tan−1u)=11+u2⋅dudx In this questionu=2x, so we get: ddx(tan−12x)=11+(2x)2⋅ddx(2x) =21+4x2 Note Ify=tan−1x, thentany=x Differentiating implicitly gets us: sec2ydydx=1, so dydx=1sec2y From trigonometry, we know that1+tan2y=sec2y sodydx=11+tan2y and we havetany=x, so we get: Fory=tan−1x, the derivative is: dydx=11+x2

Answer»

The derivative oftan−1xis11+x2(for "why", see note below)

So, applying the chain rule, we get:

ddx(tan−1u)=11+u2⋅dudx

In this questionu=2x, so we get:

ddx(tan−12x)=11+(2x)2⋅ddx(2x)

=21+4x2

Note

Ify=tan−1x, thentany=x

Differentiating implicitly gets us:

sec2ydydx=1, so

dydx=1sec2y

From trigonometry, we know that1+tan2y=sec2y

sodydx=11+tan2y

and we havetany=x, so we get:

Fory=tan−1x, the derivative is:

dydx=11+x2

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