00:01
Suppose you differentiate the function h of x, which is equal to the cube of x squared plus 3.
00:07
Now, this function, we can rewrite this as the product of x squared plus 3 and the square of x squared plus 3.
00:19
Now, the square of x squared plus 3, we can expand that.
00:24
That means we have x squared plus 3 times we have the square of x squared plus 3 times we have the square of x squared plus twice the product of x squared and three plus the square of three that gives us x squared plus three times x squared squared square which is x to the four plus six x squared plus nine and so from here we can use product rule to differentiate h now the product rule states that if a function y is written as a product of f and g, then the derivative of y, which is y prime, is equal to f g prime plus gf prime.
01:15
And so using this rule, we have h prime of x that's equal to x squared plus three times the derivative of x to the four plus six x squared plus nine, plus x to the four plus six x squared plus nine, times the derivative of x squared plus three.
01:41
And so we get x squared plus three times 4x to the third power plus 12x plus x plus x to the 4 plus 6x squared plus 9 times to x, which is equal to x squared plus 3...