differentiate $(2x^3 + 3x)(x − 2)(x + 4)$ So, I'm really stuck on this problem.
Differentiate $(2x^3 + 3x)(x − 2)(x + 4)$
This is what I come up with $10x^4+16x^3+39x^2+6x-18$.
But, the answer in the book has $16x^4$ as the leading term
Here's my work:
$(2x^3+3x)d/dx(x^2+2x-8)+(x^2+2x-8)d/dx(2x^3+3x)$
$(2x^3+3x)(2x+2)+(x^2+2x-8)(6x^2+3)$
$(4x^4+4x^3+6x^2+6x+6x^4+12x^3-48x^2+3x^2+6x-24)$
$=10x^4+16x^3+39x^2+6x-18$
 A: HINT : If $u,v,w$ are functions of $x$
Then $$\frac{d(uvw)}{dx}=uv\frac{dw}{dx}+uw\frac{dv}{dx}+vw\frac{du}{dx}$$
A: Let $y=(2x^3+3x)(x-2)(x+4)$. Then if we are not too fussy about whether this makes sense in the reals, we get
$$\ln y=\ln(2x^3+3x)+\ln(x-2)+\ln(x+4).$$
Differentiating, we get
$$\frac{1}{y}\frac{dy}{dx}=\frac{6x^2+3x}{2x^3+3x}+\frac{1}{x-2}+\frac{1}{x+4},$$
and now we know $\frac{dy}{dx}$.
In this problem, the implicit differentiation approach does not give a significicant advantage over the plain Product Rule. However, it does become useful if the problem is a little more complicated.
A: I would do it simply by expanding and then repeatedly using the Power Rule (using the Product Rule would be overkill I think):
$$
(2x^3+3x)(x-2)(x+4) = 2x^5+4x^4-13x^3+6x^2-24x.
$$
Now differentiate with respect to $x$, obtaining
$$
\frac{d}{dx}(2x^5+4x^4-13x^3+6x^2-24x) = 10x^4+16x^3-39x^2+12x-24.
$$
A: Up to $$(2x^3+3x)(2x+2)+(x^2+2x-8)(6x^2+3)$$ your work was fine but, immediately after, you have some mistakes since, developing, you are supposed to get $$(4 x^4+4 x^3+6 x^2+6 x)+(6 x^4+12 x^3-45 x^2+6 x-24)=10 x^4+16 x^3-39 x^2+12 x-24$$
