Take the derivative.

Affinity · November 23rd 2006, 08:44 AM

Take the derivative of

and

I'm just confused about where I stop for the first one, how much do I have to simplify. And for the second one, I don't know how to start it. Thanks in advance!

**galactus** · November 23rd 2006, 11:48 AM

For the second one, I know it looks scary, but break it up using the log rules and it ain't so bad.

$ln\left(\frac{x^{5}e^{4x}\sqrt{3x+1}}{1-x^{2}}\right)$

= $\left[ln(x^{5})+ln(e^{4x})+ln(\sqrt{3x+1})\right]-ln(1-x^{2})$

= $5ln(x)+4x+\frac{1}{2}ln(3x+1)-ln(1-x^{2})$

Now, you can differentiate, can't you?.

**topsquark** · November 23rd 2006, 11:56 AM

Originally Posted by Affinity

Take the derivative of

and

I'm just confused about where I stop for the first one, how much do I have to simplify. And for the second one, I don't know how to start it. Thanks in advance!

$f(x)= x \cdot ln(x + 3) - ln(9)$

$f'(x) = ln(x + 3) + \frac{x}{x + 3}$

I would leave it like this.

For the second problem, the first thing I would do would be to use the property of logarithms to split this up a bit:
$ln \left ( \frac{x^5e^{4x}\sqrt{3x+1}}{1-x^2} \right )$

= $ln(x^5) + ln(e^{4x}) + ln(\sqrt{3x+1}) - ln(1-x^2)$

= $5 ln(x) + 4x + \frac{1}{2} \cdot ln(3x + 1) - ln(1 - x^2)$

Then
$\frac{d}{dx}ln \left ( \frac{x^5e^{4x}\sqrt{3x+1}}{1-x^2} \right )$

= $\frac{5}{x} + 4 + \frac{1}{2} \cdot \frac{3}{3x+1} - \frac{1}{1 - x^2} \cdot (-2x)$

and then simplify a bit.

The brute force approach also works. It involves the chain rule, the quotient rule, the product rule, and a LOT of cancellations:
$\frac{d}{dx}ln \left ( \frac{x^5e^{4x}\sqrt{3x+1}}{1-x^2} \right )$

= $\frac{1}{\frac{x^5e^{4x}\sqrt{3x+1}}{1-x^2}}$ $\cdot ( \frac{5x^4e^{4x}\sqrt{3x+1}}{(1-x^2)}$ $+\frac{x^5 \cdot 4e^{4x} \sqrt{3x+1}}{(1-x^2)}$ $+ \frac{x^5e^{4x} \cdot \frac{3}{2\sqrt{3x+1}}}{(1 - x^2)}$ $- \frac{x^5e^{4x}\sqrt{3x+1} \cdot (-2x)}{(1-x^2)^2} )$
(Sorry. I had to split the second factor up into terms to make the LaTeX work.)

= $\frac{5}{x} + 4 + \frac{3}{2(3x + 1)} + \frac{2x}{1-x^2}$

just like before. (I highly recommend the first method, but the second one is *great* for practice!

)

-Dan

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