Class 12 Continuity And Differentiability · Long Answer3 Marksbranch_sensitive_differentiation

Class 12 Maths · Continuity And Differentiability · Long Answer #78

Differentiate y=arcsin[2^(x+1)3^x/(1+36^x)] on its differentiable real intervals.

✅ Step-by-Step Solution

Final Answer: 2·6^x ln6/(1+36^x) for x<0; its negative for x>0; no derivative at 0.

Write u=6^x>0. The arcsine argument is 2u/(1+u²)=sin(2arctan u).
If x<0 then u<1 and 2arctan u<π/2, so y=2arctan u. If x>0 then y=π−2arctan u.
Differentiate each branch: the magnitude is 2·6^x ln6/(1+36^x), positive for x<0 and negative for x>0. At 0 the one-sided derivatives are ln6 and −ln6.
Thus the derivative has the stated piecewise signs and does not exist at x=0.

[0.5 marks] Write u=6^x>0. The arcsine argument is 2u/(1+u²)=sin(2arctan u).
[1 mark] If x<0 then u<1 and 2arctan u<π/2, so y=2arctan u. If x>0 then y=π−2arctan u.
[1 mark] Differentiate each branch: the magnitude is 2·6^x ln6/(1+36^x), positive for x<0 and negative for x>0. At 0 the one-sided derivatives are ln6 and −ln6.
[0.5 marks] Thus the derivative has the stated piecewise signs and does not exist at x=0.

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