# Thread: stuck on a calculus problem from online homework(logarithmic differentiation)

1. ## stuck on a calculus problem from online homework(logarithmic differentiation)

I just registered and am excited to join the community. My problem is: . I feel like I am very close but have overlooked some things. I am aware that derivative of ln is 1/x but I can't seem to get the final answer right. Also, one that has been bugging me for a few days is . For this one I have gotten stuff like 1/xsqrt(x^2-5)*sqrt(x^2-5)+1/(2(sqrt(X^2-5))) *2x^2. Any help would be much appreciated.

2. Originally Posted by nvr20
I just registered and am excited to join the community. My problem is: . I feel like I am very close but have overlooked some things. I am aware that derivative of ln is 1/x but I can't seem to get the final answer right. Also, one that has been bugging me for a few days is . For this one I have gotten stuff like 1/xsqrt(x^2-5)*sqrt(x^2-5)+1/(2(sqrt(X^2-5))) *2x^2. Any help would be much appreciated.
$g = \ln{(x\sqrt{x^2 - 5})}$.

Let $u = x\sqrt{x^2 - 5}$ so that $g = \ln{u}$.

$\frac{dg}{du} = \frac{1}{u} = \frac{1}{x\sqrt{x^2 - 5}}$.

$\frac{du}{dx} = x\,\frac{d}{dx}(\sqrt{x^2 - 5}) + \sqrt{x^2 - 5}\,\frac{d}{dx}(x)$

$= x\cdot \frac{x}{\sqrt{x^2 - 5}} + \sqrt{x^2 - 5}\cdot 1$

$= \frac{x^2}{\sqrt{x^2 - 5}} + \sqrt{x^2 - 5}$

$= \frac{x^2\sqrt{x^2 - 5}}{x^2 - 5} + \frac{(x^2 - 5)\sqrt{x^2 - 5}}{x^2 - 5}$

$= \frac{(2x^2 - 5)\sqrt{x^2 - 5}}{x^2 - 5}$.

Therefore $\frac{dg}{dx} = \frac{dg}{du}\,\frac{du}{dx}$

$= \left(\frac{1}{x\sqrt{x^2 - 5}}\right)\left(\frac{(2x^2 - 5)\sqrt{x^2 - 5}}{x^2 - 5}\right)$

$= \frac{2x^2 - 5}{x(x^2 - 5)}$.

3. Originally Posted by nvr20
I just registered and am excited to join the community. My problem is: . I feel like I am very close but have overlooked some things. I am aware that derivative of ln is 1/x but I can't seem to get the final answer right. Also, one that has been bugging me for a few days is . For this one I have gotten stuff like 1/xsqrt(x^2-5)*sqrt(x^2-5)+1/(2(sqrt(X^2-5))) *2x^2. Any help would be much appreciated.
$H(z) = \ln{\left(\sqrt{\frac{1 - z^2}{1 + z^2}}\right)}$

$= \ln{\left(\frac{1 - z^2}{1 + z^2}\right)^{\frac{1}{2}}}$

$= \frac{1}{2}\ln{\left(\frac{1 - z^2}{1 + z^2}\right)}$.

Now let $u = \frac{1 - z^2}{1 + z^2}$ so that $H = \frac{1}{2}\ln{u}$.

$\frac{dH}{du} = \frac{1}{2u} = \frac{1}{2\left(\frac{1 - z^2}{1 + z^2}\right)} = \frac{1 + z^2}{2(1 - z^2)}$.

$\frac{du}{dx} = \frac{(1 + z^2)\frac{d}{dx}(1 - z^2) - (1 - z^2)\frac{d}{dx}(1 + z^2)}{(1 + z^2)^2}$

$= \frac{-2z(1 + z^2) - 2z(1 - z^2)}{(1 + z^2)^2}$

$= \frac{-2z - 2z^3 - 2z + 2z^3}{(1 + z^2)^2}$

$= \frac{-4z}{(1 + z^2)^2}$.

So $\frac{dH}{dx} = \frac{dH}{du}\,\frac{du}{dx}$

$= \left( \frac{1 + z^2}{2(1 - z^2)}\right)\left(\frac{-4z}{(1 + z^2)^2}\right)$

$= -\frac{2z}{(1 - z^2)(1 + z^2)}$.

4. Thanks a lot! Where I went wrong was the differentiation of \sqrt{x^2 - 5}. Makes much sense now.