Find first derivatives

Question

log√1+sinx

       1-sinx

Answer 1

first, let me assume your equations are: LOG(√(1+SINX) and LOG(√(1-SINX)
and that the LOG is the Logarithm with base 10 ?? or natural Logarithm LN ?? I'll answer both ways.

1) F(X) = LOG10(√(1+SINX)
THEN F(X) = LOG10(1+SINX)1/2 and using rules of logarithms
F(X) = (1/2)LOG10(1+SINX)

NOW, THE DERIVATIVE F'(X) = (1/2) (1/LN(10)) * (1/(1+SINX)) * COSX or
F'(X) = COSX / [2*LN(10)*(1+SINX)]

Note: if equation 1 was using LN instead of LOG, the answer would be
F'(X) = COSX / [2*(1+SINX)]


2) F(X) = LOG10(√(1-SINX)

Similarly, Then
F'(X) = - COSX / [2*LN(10)*(1-SINX)] (note the (-) sign)

Note: if equation 2 was using LN instead of LOG, the answer would be
F'(X) = - COSX / [2*(1-SINX)]