Compute the differential \(dy\):
\[ y = (2x+sinx)^2 \] Because this can be thought of as a compound function, we can use chain rule to solve the derivative.
Let us say that \(y = f(x)\), so that \(f(x)=(2x+sinx)^2\). We can set the function being squared equal to \(g(x)\), such that \(g(x)=2x+sinx\).
Per the chain rule, when the derivative of \(g(x)^2\) is \(2g(x)g′(x)\). Therefore, if we find \(g'(x)\) (and already knowing \(g(x)\)) we can solve for \(f'(x)\), or \(y'\). And because \(g(x)\) is a polynomial, the derivative can be treated as a sum of derivatives.
\[ g'(x) = \frac{d}{dx}(2x+sinx) \\ = \frac{d}{dx}(2x)+\frac{d}{dx}(sinx) \\ = 2 + cosx \] Now, plugging \(g(x)\) and \(g'(x)\) into our equation for \(f'(x)\):
\[ f'(x) = 2g(x)g′(x) \\ = 2(2x+sinx)(2+cosx) \\ \] While this function can be manipulated in various ways, this may be the simplest and cleanest way to present it.