Using the product rule and not the quotient rule, what is the derivative with respect to a of the following?

Question

F(a) = sin(a/2 + d/2) / sin(a/2)

Please use the product rule only (not the quotient rule) to find the derivative of F(a) with respect to a.

Thanks.

Answer 1

F(a) = sin(a/2 + d/2) / sin(a/2)

Given that we want to use the product rule instead of the quotient rule, let's rewrite this:

F(a) = sin(a/2 + d/2) * [1/sin(a/2)]

The definition of cosecant is one over sine, so let's convert this to csc(a/2).

F(a) = sin(a/2 + d/2) (csc(a/2))

Now, use the product rule. Remember that the derivative of csc(x) = -csc(x)cot(x). In this case, we apply the chain rule as well as the product rule.

F'(a) = cos(a/2 + d/2)(1/2) csc(a/2) + sin(a/2 + d/2)[-csc(a/2)cot(a/2)] (1/2)

F'(a) = (1/2)cos(a/2 + d/2)csc(a/2) - (1/2)sin(a/2 + d/2)csc(a/2)cot(a/2)

Answer 2

Rewrite sin(a/2 + d/2) / sin(a/2) as...

sin(a/2 + d/2) * (sin a/2)^-1

Then use the product rule (and chain rule) as appropriate.

Answer 3

f(a) = sin(a/2 + d/2)*(1/sin(a/2))
f'(a) = cos(a/2 + d/2)*(1/2)*(1/sin(a/2)) + sin(a/2 + d/2) * (-1/sin^2(a/2)) * (cos(a/2)) * (1/2)

= cos(a/2 + d/2)/(2sin(a/2)) - sin(a/2 + d/2)cos(a/2)/(2sin^2(a/2))

= [ cos(a/2 + d/2)*sin(a/2) - sin(a/2 + d/2)cos(a/2)]/[2sin^2(a/2)]

using the difference of angles formula for sine function:

= sin(d/2)/(2sin^2(a/2))

Hope I didn't make any mistakes....

Answer 4

a million) y=ln(lnx) y '=(a million/lnx)(a million/x)=> y '=a million/[xln(x)]...(use the chain rule) 2) y=sqrt[x(x+a million)] y '=a million/[2sqrt(x(x+a million)](2x+a million)=> y '=(2x+a million)/[2sqrt(x(x+a million)] 3) y=sinAsqrt(A+3) y '=cosAsqrt(A+3)+sinA/[2sqrt(A+3)] y '=[2cosA(A+3)+sinA]/[2sqrt(A+3)] 4) y=(A+5)/(AcosA) y '=[AcosA-(A+5)(cosA-AsinA)]/ ......(AcosA)^2=> y '=(A+5)sinA/(Acos^2A)