she thought she would get points of infliction!
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the second derivative at an inflectiion point is zero
When the first derivative of the function is equal to zero and the second derivative is positive.
You are supposed to use the chain rule for this. First step: derivative of root of sin2x is (1 / (2 root of sin 2x)) times the derivative of sin 2x. Second step: derivative of sin 2x is cos 2x times the derivative of 2x. Third step: derivative of 2x is 2. Finally, you need to multiply all the parts together.
(x3)'=3x2(x3)''=(3x2)'=6x
When you solve for the 2nd derivative, you are determining whether the function is concave up/down. If you calculated that the 2nd derivative is negative, the function is concave down, which means you have a relative/absolute maximum, given that the 1st derivative equals 0. To understand why this is, think about the definition of the 2nd derivative. It is a measure of the rate of change of the gradient. At a maximum, the gradient starts positive, becomes 0 at the maximum itself and then becomes negative, so it is decreasing. If the gradient is going down, then its rate of change, the 2nd derivative, must be negative.