Question

d/dx (a +bsinx/a-bsinx)

Answer 1

=d/dx(a+bsinx÷a-bsinx)
By using Quotient rule;
= (a-bsinx)(0+bcosx)- (a+bsinx)(0-bcosx)
= (a-bsinx)(bcosx)- (a+bsinx)(bcosx)
= (bcosx)[a-bsinx+a+bsinx]
=2abcosx.
hence found
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Answer 2

Given :

f(x) = a + bsinx/a - bsinx

To find :

d (a + bsinx/a - bsinx)/ dx

Solution :

Let u = a + bsinx and v = a - bsinx

• Using quotient rule for differentiating above function

Given by,

d f(x) /dx = u'×v - v'×u / u^2

d f(x) /dx = [d(a + bsinx)/dx]×(a - bsinx) - [d(a - bsinx)/dx]×(a + bsinx) / (a - bsinx)^2

= bcosx(a - bsinx) - bcosx(a + bsinx) / (a - bsinx)^2

= abcosx - b^2cosx*sinx - abcosx - b^2cosx*sinx / (a - bsinx)^2

= -b^2cosx*sinx / (a - bsinx)^2

Hence, final answer is

d(a +bsinx/a-bsinx) / dx = -b^2cosx*sinx / (a - bsinx)^2