Question

Not really understanding how the derivative of y=sin^2(x) is 2cosx*sinx? I know we use the chain rule, but steps would help.

Answer 1

`y=sin^2(x)\iff y=\left[ sin(x) \right]^2\impliedby \textit{pretty much} \\ \quad \\ \textit{now, using the chain-rule} \\ \quad \\ \cfrac{dy}{dx}=2[sin(x)]^1\cdot cos(x)\\ \quad \ \\ \quad \\ \textit{the cos(x), is the derivative of the inner function}`

when it comes to trigonometric functions and exponentials,
the exponent usually as a matter of notation, goes with the
name of the trig function, and then you put the argument or
parameters ahemm... that's just a notational issue

now, if you were to do as above, so simply group the whole
function in parentheses or brackets, and THEN apply the
exponential, the expression is exactly the same,
but the grouping makes it more conspicuous what the exponent,
is comprising