anti-derivatives

Question

Compute the derivative d/dx sin2x. Using this, find (with an explanation) an antiderivative for cos(2x)

Compute the derivative d/dx e^{-x^2} Using this, find (with an explanation) an antiderivative for xe^{-x^2}

Compute the derivatives d/dx ex and d/dx e^x and d/dx(xe^x) Using these, find (with an explanation) an antiderivative for xe^x

Answer 1

Part 1

The general derivative for sin(x) is cos(x), and by using the chain rule:

d/dx sin(2x) = (2x)' ⋅ cos(2x) = 2cos(2x)

Hence, if we were to find the antiderivative of cos(2x), which is half of what we had above, then it would be (1/2)sin(2x) because the properties of the chain rule will help cancel out the fraction as shown previously.

Part 2

The general derivative for e^x is e^x, or itself, and by using the chain rule:

d/dx e^{-x^2} = (-x²)' ⋅ e^{-x^2} = -2xe^{-x^2}

Hence, if we were to find the antiderivative of xe^{-x^2}, this is -1/2 times the derivative we had originally, so the antiderivative would be (-1/2)e^{-x^2} because the properties of the chain rule will help cancel out the fraction as shown previously.

Part 3

The derivative of xe^x can be calculated by the product rule:

d/dx xe^x = (x)'e^x + x(e^x)' = e^x + xe^x = e^x(x+1)

As you may notice, the factor x is increased to x+1 to get the derivative, so the antiderivative of xe^x would be e^x(x-1) following this similar logic.

Think of the power rule with a function like x². Its antiderivative would be x²⁺¹/(2+1) = x³/3, and then differentiating that would result in 3(1/3)x^3-1 = x².

Hope this helped!