# can someone please just double check

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• Jul 1st 2008, 09:28 PM
10164064
can someone please just double check
I just want to make sure I am doing this correctly

d/dx(e^5x + 1) = 5e^5x

and

find d2/dx2 (lnx)^3 = 6(lnx)
• Jul 1st 2008, 09:43 PM
Mathstud28
Quote:

Originally Posted by 10164064
I just want to make sure I am doing this correctly

d/dx(e^5x + 1) = 5e^5x

and

find d2/dx2 (lnx)^3 = 6(lnx)

Second one is wrong

Remember

$\displaystyle \frac{d}{dx}\bigg[f(g(x))\bigg]=f'(g(x))\cdot{g'(x)}$
• Jul 1st 2008, 09:47 PM
Chris L T521
Quote:

Originally Posted by 10164064
I just want to make sure I am doing this correctly

d/dx(e^5x + 1) = 5e^5x

and

find d2/dx2 (lnx)^3 = 6(lnx)

The first one is correct.

However, the second one is missing something...

$\displaystyle \frac{d^2}{dx^2}(\ln(x))^3$

You need to apply the chain rule here...

$\displaystyle \frac{d}{dx}(\ln(x))^3=3(\ln(x))^2\cdot\frac{1}{x} =\frac{3(\ln(x))^2}{x}$

$\displaystyle \frac{d}{dx}\frac{3(\ln(x))^2}{x}=\frac{6\ln(x)-3(\ln(x))^2}{x^2}$

Hope this makes sense! :D

--Chris
• Jul 1st 2008, 09:51 PM
10164064
its to the second derivative right?
so is it
{6/x(lnx)}-{3(lnx)^2/x^2}
• Jul 1st 2008, 09:54 PM
10164064
Ok thanx for the clarification Chris.