Check out some similar questions!

dths1980 · Jun 20, 2008, 12:23 PM

f(x)=x^3ln(2x^4+7)

please help me with this problem, thanks.

galactus · Jun 20, 2008, 12:35 PM

This requires the product and chain rules.

$x^{3}\cdot\frac{8x^{3}}{2x^{4}+7}+3x^{2}ln(2x^{4}+ 7)$

See what I done? The product rule is $f(x)g'(x)+g(x)f'(x)$

The derivative of $ln(2x^{4}+7)$ requires the chain rule. That is essentially the

derivative of the inside times derivatives of outside.

Since the derivative of $ln(u)=\frac{1}{u}$ , then we have

$\frac{1}{2x^{4}+7}$ . But we multiply that by the derivative of the inside,

which is $\frac{d}{dx}[2x^{4}+7]=8x^{3}$

See? I will let you tidy it up.

Unknown008 · Jun 21, 2008, 02:18 AM

Yes, that's it. Too bad, I wasn't here yesterday to answer before you!