Question

Differentiate Sin inverse of 2 X upon oneplus x square with respect to Cos inverse of 1 - x square upon oneplus x square

Answer 1

here is your answer hope it will help you

Answer 2

Question: Differentiate sin⁻¹{2x/(1 + x²)} with respect to cos⁻¹{(1 - x²)/(1 + x²)}.

Step-by-step explanation:

Let x = tanθ

Then m = sin⁻¹ {2x/(1 + x²)}

= sin⁻¹ {2 tanθ/(1 + tan²θ)}

= sin⁻¹ (2 tanθ/sec²θ)

= sin⁻¹ (2 sinθ cosθ)

= sin⁻¹ (sin2θ)

= 2θ

and n = cos⁻¹ {(1 - x²)/(1 + x²)}

= cos⁻¹ {(1 - tan²θ)/(1 + tan²θ)}

= cos⁻¹ {(1 - tan²θ)/sec²θ}

= cos⁻¹ (cos²θ - sin²θ)

= cos⁻¹ (cos2θ)

= 2θ

We have to find dm/dn

Now, dm/dθ = 2 & dn/dθ = 2

So dm/dn = (dm/dθ) * (dθ/dn)

= 2 * 1/2

= 1

∴ d/d[cos⁻¹{(1 - x²)/(1 + x²)}] [sin⁻¹{2x/(1 + x²)}] = 1