Question

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

Answer 1

\[y=-122x *e ^{-2x ^{3}}+(-61x ^{2})*(e ^{-2x ^{3}})*(-6x^2)\]

Answer 2

y' *

Answer 3

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.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 \[\frac{ dy }{ dx }= 366x^4\exp(−2x^3)−122xexp(−2x^3)\]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:)