If y=(a^(Sinx)) where a and k are given constants, find dy/dx in terms of a and x

Here we have to differentiate a constant raised to the power of a variable. To make it easier, let u=sinx and so our function can now be treated as y=a^u. Remembering that A = e^(LnA), a^u = e^(Ln(a^u)). Using our log laws, we know that Ln(a^u) = uLn(a). This is now much easier to approach. Since a is a constant, Ln(a) is also a constant. Therefore the derivative (with respect to u) of e^(uLn(a)) is simply Ln(a)e^(uLn(a)). Remembering that a^u = e^(Ln(a^u)), we can rewrite this as Ln(a)a^u.
So we have worked out dy/du. Going back to our u=sinx, we know that du/dx=cosx.The question asks for dy/dx. Using the chain rule, we know that dy/dx = (dy/du)(du/dx)
So dy/dx = Cos(x)Ln(a)(a^sinx)