Using the sum, chain and product rules, differentiate the function f(x) = x^n +x^3 * sin(1/[3x])

The following function consists of the sum of two variables x^n and x^3 * sin(1/[3x]). Let's break the differentiation down term by term and differentiate xⁿ first since it's the easier term. Recalling the sum rule of differentiation; d[xⁿ]/dx = nx^(n-1). So what we did here is some standard differentiation, where we multiply the function by the exponent, and subtract a one from its exponent. For the 2^nd term; x^3 * sin(1/3x) this is a product of two functions that cannot be simplified further, and therefore we need to use the product rule to differentiate it. Recalling the derivative of the product of two functions; d(f.g)/dx = f.dg/dx + g.df/dx , where f and g are two functions of x. So, we keep one of the functions "constant" while we differentiate the other function, and vice versa for the second function, and then we sum them up to give us our final derivative. But how do we differentiate the sin(1/3x)? The derivative of sin(x) is cos(x). To differentiate sin(1/3x) we need to use the chain rule which is; df(g(x))/dx = dg/dx * df(g(x))/dx, where in our case g(x) = 1/3x and f(g(x)) is sin(1/[3x]). Therefore, by performing the operation we find that; d[sin(1/3x)]/dx = -1/(3x^2)cos(1/3x). So using the product rule we find that; d[x^3 sin(1/3x)]/dx = 3x²*sin(1/3x)-[x /3]cos(1/3x). And finally our answer to the original function is; df(x)/dx = nx^(n-1)  + 3x²*sin(1/3x)-[x /3]*cos(1/3x)