Question

differentiate: y=-61x^2e^(-2x^3)

Answer 1

Okay, this one is product rule, then chain rule on the exponential. Start by letting: \[u = -61x^2 \]\[v= \exp(-2x^3)\] (sorry to use exp instead of e^, just looks a little clearer on the formula).

From there we see that \[y=uv\]therefore by product rule: \[\frac{ dy }{ dx } = u \frac{ dv }{ dx } + v \frac{ du }{ dx }\] So we need to differentiate the u and v we defined earlier. u is easy, however v is a touch more difficult. We see that \[\frac{ du }{ dx } = -122x\] and by the rules of exponents, \[\frac{ dv }{ dx } = -6x^2\exp(-2x^3)\] since when differentiating an exponential function, you differentiate the bit at the top, then just multiply that by the original thing. ie. \[\frac{ d }{ dx }(\exp(ax)) = aexp(ax)\] Now by just substituing all the terms we have found, we can see that \[\frac{ dy }{ dx } = (-61x^2)(-6x^2\exp(-2x^3)) + (\exp(-2x^3))(-122x)\] expand this out to get \[\frac{ dy }{ dx } = 366x^4\exp(-2x^3) -122x \exp(-2x^3)\]from here you can tidy it up a little bit to get \[\frac{ dy }{ dx }= 122x \exp(-2x^3)(3x^3 - 1)\] Hope that this helped:)