Integration by parts looks like a promising first step.cooper607 said:hi , i need some help in this integral
e^(-bx^2) d^2/dx^2 (e^(-bx^2)) dx from - to + infinityI tried differentiating e^(-bx^2) twice and it came up weird , is there any other way to do it ?
The general formula for differentiating e^(-bx^2) is d/dx(e^(-bx^2)) = -2bxe^(-bx^2).
To use the power rule for differentiating e^(-bx^2), you simply multiply the function by the derivative of the exponent, which in this case is -2bx. So, the derivative is -2bxe^(-bx^2).
The chain rule for differentiating e^(-bx^2) involves taking the derivative of the outer function, e^(-bx^2), and multiplying it by the derivative of the inner function, which is -bx. So, the final result is -2bxe^(-bx^2).
The natural logarithm is used when differentiating e^(-bx^2) because it allows us to simplify the expression by bringing down the exponent as a coefficient. The derivative then becomes -2bx.
There are no special cases or exceptions when differentiating e^(-bx^2). The power rule and chain rule can be used to find the derivative in all cases. However, it is important to remember that the derivative of e^x is e^x and does not change when differentiating e^(-bx^2).