Chain Rule for Derivatives. Example Problem 2
Let’s take the derivative of x cubed times the sine of x squared. To do this problem, we will first look at it globally, considering both functions as a whole. The derivative is fg prime plus f prime g. So let’s color code this: fand g are blue; x cubed is yellow; sin x squared is orange; and their product is purple The chain rule allows us to take the derivative of composite functions. In this case, it’s a composite function because there’s something on the inside of some parentheses that is not simply an x. A composite function: cosign times the square root of x squared plus 2x. So what we do is we take the derivative of cosign, which is cosine × squared. The inside component will stay the same, and multiply it by the derivative of the inside, which is 2x. So there you have the derivative of sine of x squared. That’s 3x squared times sine × squared plus write down the derivative of f3x squared and multiply that by g(sin x), or sine › squared. And then we can clean everything up 2x cubed times 2x; well, that’s 4x cubed plus 3x squared times sin x squared becomes 4x cubed plus 3x cubed times sin x cubed. d/dx(x^3 *sin(x^2)) fg’+f’g | d/dx sin(x^2) x^3 *cos(x^2)*2x+3x^2 *sin(x^2) | cos(x^2)*2x 2x^4 *cos(x^2)+3x^2 *sin(x^2) | ______________
