Differentiating $y=x^6e^{-4x^3}$ Can some one explain, how to solve this derivative.
I'm total beginner.
Would be preferable if some one explain step-by-step.
$$y=x^6e^{-4x^3}$$
 A: Breaking everything, let me start by remembering a couple of rules, if $f, g$ are functions:
$$(fg)' = f'g + fg' \\ [f(g)]' = f'(g)\cdot g$$
those are just the rules of the derivatives of the product and of the composition.
Let $f(x) = x^6$ and $g = e^{-4x^3}$. What you have is
$$y = fg$$
Therefore differentiating it gives
$$y' = (fg)' = f'g + fg' = (x^6)'\cdot e^{-4x^3} + x^6\cdot (e^{-4x^3})'$$
Let us now write $h(x) = e^x$, $j(x) = -4x^3$. Notice that the rightmost part is
$$x^6\cdot (e^{-4x^3})' = x^6\cdot [h(j(x))]'$$
Using the chain rule,
$$x^6\cdot [h(j(x))]' = x^6\cdot j(x)'\cdot h'(j(x)) = x^6 \cdot (-4x^3)' \cdot h'(-4x^3)$$
Now notice that $(e^x)' = e^x$ and therefore $h' = h$:
$$x^6 \cdot (-4x^3)' \cdot h'(-4x^3) = x^6 \cdot (-4x^3)' \cdot e^{-4x^3}$$
Assuming you know the rule $(ax^k)' = (kax^{k-1})$, we can use it to finish everything. We had $y' = (x^6)'\cdot e^{-4x^3} + x^6\cdot (e^{-4x^3})'$. We substitute the rightmost part by what we got:
$$y' = (x^6)'\cdot e^{-4x^3} + x^6\cdot (-4x^3)' \cdot e^{-4x^3}$$
and apply the rule I just told you about: $(x^6)' = 6x^5, (-4x^3)' = -12x^2$ giving
$$y' = 6x^5\cdot e^{-4x^3} + x^6\cdot -12x^2 \cdot e^{-4x^3} = 6x^5\cdot e^{-4x^3} - 12x^8\cdot e^{-4x^3}$$
A: you must use the product rule, this means $$u'v+uv'$$
and we have $$u=x^6$$ thus $$u'=6x^5$$, now it is $$v=e^{-4x^3}$$ and we need the chaine rule: $$v'=e^{-4x^3}\cdot(-12x^2)$$
now you can build the derivative.
A: When you face expressions which only contain products, quotients and powers, logarithmic differentiation makes life easier $$y=x^6e^{-4x^3}\implies \log(y)=6\log(x)-4x^3$$ Now, differentiation $$\frac {y'}y=\frac 6 x-12 x^2=\frac 6 x(1-2x^3)$$ $$y'=y \frac 6 x(1-2x^3)=x^6e^{-4x^3}\frac 6 x(1-2x^3)=6x^5(1-2x^3)e^{-4x^3}$$ 
