Find the derivative of StartFraction d Over dx EndFraction Integral from 0 to x cubed e Superscript negative t Baseline font size decreased by 3 dt

a. by evaluating the integral and differentiating the result.
b. by differentiating the integral directly.

(a) e ^ -3x (b)e^-3x

Step-by-step explanation:

I suggest the equation is:

d/dx[integral (e^-3t) dt

First we integrate e^-3tdt

Integral(e ^ -3t dt) as shown in attachment and then we differentiate the result as shown in the attachment.

(b) to differentiate the integral let x = t, and substitute into the expression.

Therefore dx = dt

Hence, d/dx[integral (e ^-3x dx)] = e^-3x

(a)  3x²e^(-x³)

(b) -3x²e^(-x³)

Step-by-step explanation:

(a)  The integral was evaluated at t = 0 to t = x³

Then the result is differentiated to obtain the result.

(b) The integral is differentiates directly, using the properties of differentiation, the chain rule precisely.

The result obtained in (a) turned out to be the negative of the result obtained in (b)

CHECK ATTACHMENT FOR THE WORKINGS.

me! what did you put for this question?

step-by-step explanation: